Jennifer went to a farm and bought tomatoes
for $2 per pound and garlic for $3 per pound.
If Jennifer spent $72 on tomatoes and garlic,
how many pounds can she buy if she buys
only garlic?

The probability that a person wins the game is 32.1%

Based on the information given, the following can be depicted. It should be noted that there are 6 sides as well as 4 cards.

Therefore, the numbers on the dice i.e from 1 - 6 will be represented 4 times each. This gives a total of (4 × 6) = 24. There are also 4 cards. The total in sample space will now be:

= 24 + 4 = 28

The frequency table will be such that 28 or more has a relative frequency of 9. Therefore, the probability that a person wins the game will be:

= 9/28 = 32.1%

When you win 25% of the time, this illustrates that the number of products picked will be:

= 25% × 28

= 7 products.

The probability of participants achieving a winning score of 36 or higher in four consecutive attempts will be:

= 1/6⁴

= 1/1296

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The value of x is the length GH, which is determined as 6.

The value of x is the length GH and can be calculated from Pythagoras theorem.

|OG|² = |OS|² + |SG|²

|OG|² = 9² + 12²

|OG|² =  225

|OG| = √225

|OG| = 15

x + H = |OG|

where;

x + 9 = 15

x = 15 - 9

x = 6

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wth is that wdym ???

Using an exponential function, it would take 824 minutes before the amount of medication in Aldi's bloodstream is effectively 0.

A decaying exponential function is given as;

[tex]A(t) = A(0)e^-kt[/tex]

Where A = initial value

k = decay constant

We have the half-life of amoxicillin as 62minutes, to determine k, we get

[tex]A(62) = 0. 5(0)[/tex]

[tex]A(62) = 0.5A(0)[/tex]

[tex]0. 5A(0) = A(0) e^{-62k}[/tex]

Solve the exponential function thus

[tex]e^{-62k} = 0. 5[/tex]

㏑ [tex]e^{-62k}[/tex] = ㏑ 0. 5

- 62k = ㏑ 0.5

Make k subject of formula

k = -㏑[tex]\frac{0. 5}{62}[/tex]

k = 0. 011179

The equation is given by

[tex]A(t) = A(0)e^-0.011179[/tex]

We have

[tex]0. 0001A(0) = A(0)e^-0.01179[/tex]

[tex]e^-0.011179t = 0. 0001[/tex]

㏑ [tex]e^-0.011179[/tex] = ㏑ 0.0001

[tex]-0.011179t = In 0.0001[/tex]

t = [tex]-\frac{In 0. 0001}{0. 011179}[/tex]

t = [tex]824[/tex]

Therefore, it would take 824 minutes before the amount of medication in Aldi's bloodstream is effectively 0.

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The scale factor of the dilation shown in the image is: C. 4.

Scale factor = New dimension/corresponding old dimension.

From the image given, we have:

Scale factor = OA'/OA

OA' = 3 + 9 = 12 units

OA = 3 units

Scale factor of dilation = 12 units/3 units

Scale factor of dilation = 4

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C. 4

Next one is 6 if you have the option on edge

What you can do that can be set up as a two- step equation is to follow the form which is  [tex]ax + b = c[/tex] and should be noted that  a, b, c are real numbers.

Two- step equation can be solved by following these

Two step equations serves as  algebraic equations which can be solved in exactly two steps to get the  final value of the variable .

Example of this is [tex]2x + 3 = 7.[/tex]

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0 ≤ x ≤ 60

0 ≤ y ≤ 40

Definition of Domain and Range

The set of  all the possible input values for which the provided function is defined is termed as the domain of that function.

The set of all the possible values that a given function can generate as an output is termed as the range of that function.

Forming the Relation

It is given that,

Thus, the relation formed is given by,

[tex]2x + 3y \leq 120[/tex]

Domain and Range of the Given Relation

We have,

[tex]2x + 3y \leq 120[/tex]

⇒ [tex]2x \leq 120[/tex]

⇒ [tex]x\leq60[/tex]

Thus, the domain is given by,

⇒ [tex]0\leq x\leq 60[/tex]

Similarly,

[tex]3y\leq 120[/tex]

⇒ [tex]y\leq 40[/tex]

Thus, the range is given by,

⇒ [tex]0\leq y\leq 40[/tex]

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Answer:

-14 or (6,20)

Question 4

1) [tex]\overline{BD}[/tex] bisects [tex]\angle ABC[/tex], [tex]\overline{EF} \perp \overline{AB}[/tex], and [tex]\overline{EG} \perp \overline{BC}[/tex] (given)

2) [tex]\angle FBE \cong \angle GBE[/tex] (an angle bisector splits an angle into two congruent parts)

3) [tex]\angle BFE[/tex] and [tex]\angle BGE[/tex] are right angles (perpendicular lines form right angles)

4) [tex]\triangle BFE[/tex] and [tex]\triangle BGE[/tex] are right triangles (a triangle with a right angle is a right triangle)

5) [tex]\overline{BE} \cong \overline{BE}[/tex] (reflexive property)

6) [tex]\triangle BFE \cong \triangle BGE[/tex] (HA)

Question 5

1) [tex]\angle AXO[/tex] and [tex]\angle BYO[/tex] are right angles, [tex]\angle A \cong \angle B[/tex], [tex]O[/tex] is the midpoint of [tex]\overline{AB}[/tex] (given)

2) [tex]\triangle AXO[/tex] and [tex]\triangle BYO[/tex] are right triangles (a triangle with a right angle is a right triangle)

3) [tex]\overline{AO} \cong \overline{OB}[/tex] (a midpoint splits a segment into two congruent parts)

4) [tex]\triangle AXO \cong \triangle BYO[/tex] (HA)

5) [tex]\overline{OX} \cong \overline{OY}[/tex] (CPCTC)

Question 6

1) [tex]\angle B[/tex] and [tex]\angle D[/tex] are right angles, [tex]\overline{AC}[/tex] bisects [tex]\angle BAD[/tex] (given)

2) [tex]\overline{AC} \cong \overline{AC}[/tex] (reflexive property)

3) [tex]\angle BAC \cong \angle CAD[/tex] (an angle bisector splits an angle into two congruent parts)

4) [tex]\triangle BAC[/tex] and [tex]\triangle CAD[/tex] are right triangles (a triangle with a right angle is a right triangle)

5) [tex]\triangle BAC \cong \triangle DCA[/tex] (HA)

6) [tex]\angle BCA \cong \angle DCA[/tex] (CPCTC)

7) [tex]\overline{CA}[/tex] bisects [tex]\angle ACD[/tex] (if a segment splits an angle into two congruent parts, it is an angle bisector)

Question 7

1) [tex]\angle B[/tex] and [tex]\angle C[/tex] are right angles, [tex]\angle 4 \cong \angle 1[/tex] (given)

2) [tex]\triangle BAD[/tex] and [tex]\triangle CAD[/tex] are right triangles (definition of a right triangle)

3) [tex]\angle 1 \cong \angle 3[/tex] (vertical angles are congruent)

4) [tex]\angle 4 \cong \angle 3[/tex] (transitive property of congruence)

5) [tex]\overline{AD} \cong \overline{AD}[/tex] (reflexive property)

6) [tex]\therefore \triangle BAD \cong \triangle CAD[/tex] (HA theorem)

7) [tex]\angle BDA \cong \angle CDA[/tex] (CPCTC)

8) [tex]\therefore \vec{DA}[/tex] bisects [tex]\angle BDC[/tex] (definition of bisector of an angle)

Answer:

1

Step-by-step explanation:

 12+[2-(4 × a²)] ÷ 7+b

= 12+[2-(4 × 5²)] ÷ 7+ 1

= 12+[2-(4 × 25)] ÷ 7+ 1

= 12 + (2 - 100)÷ 7 + 1

= 12 - 98 ÷ 7 + 1

= 12 - 14 + 1

= 1

Answer:

-1

Step-by-step explanation:

The speed of Sali is 18.75 km/hr.

Given,

Total distance travelled by Jane and Sali =63 km.

The total time is taken by Jane=3.5 hours.

Sali started to cycle 4 minutes after Jane started to cycle [tex]=\frac{4}{60}=\frac{1}{15}[/tex] hours.

The formula to find speed is [tex]Speed=\frac{Distance}{Time} \ km/hr[/tex].

The speed of Jane[tex]=\frac{63}{3.5}=18 \ km/hr[/tex].

Let the speed of Sali be [tex]x \ km/hr[/tex].

Sali caught up with Jane when they had both cycled 30 km.

Thus, [tex]\frac{30}{18}-\frac{30}{x}=\frac{1}{15}[/tex]

[tex]\implies \ \frac{30x-540}{18x} =\frac{1}{15}[/tex]

[tex]\implies \ 15(30x-540)=18x[/tex]

[tex]\implies \ 5(30x-540)=6x[/tex]

[tex]\implies \ 150x-2700=6x[/tex]

[tex]\implies \ 144x=2700\implies \ x=18.75[/tex]

Therefore, the speed of Sali is 18.75 km/hr.

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The margin of error for a 95% confidence interval for the population proportion is 0.0815.

Given sample size=60 and confidence interval of 95%.

We have to calculate margin of error for a confidence interval of 95%.

Margin of error is the difference between actual values and calculated values. Margin of error is a part of z test.

We have been given sample size=60.

8 people have received help from their parents from the sample.

p=8/60=0.13

which is sample proportion.

z=1-0.13

=0.87

To calculate standard error=[tex]\sqrt{p*z/n}[/tex]

=[tex]\sqrt{0.13*0.8/60}[/tex]

=0.0416

at 95% confidence interval

z(α/2)=1.96

therefore margin of error=1.96*0.0416

=0.081536

=0.0815

Hence the margin of error for 95% confidence interval is 0.0815

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Since it has a negative discriminant, the quadratic equation that does not have a real solution is given by:

−4x2 − 4x − 6 = 0.

A quadratic equation is modeled by:

[tex]y = ax^2 + bx + c[/tex]

The discriminant is:

[tex]\Delta = b^2 - 4ac[/tex]

The solutions are as follows:

We want [tex]\Delta < 0[/tex], hence, for equation -4x² - 4x - 6, we have that a = -4, b = -4, c = -6, hence:

[tex]\Delta = (-4)^2 - 4(-4)(-6) = 16 - 96 = -80[/tex]

Negative discriminant, hence it does not have a real solution.

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Answer:

The answer above is correct. The answer was confirmed by a teacher, and was marked as correct on the test

Step-by-step explanation:

Answer:

the third option

Step-by-step explanation:

Our equation is

[tex] {y}^{2} = 36(x - 10)[/tex]

Equation of a vertical parabola is

[tex]( {y - k)}^{2} = 4p(x - h)[/tex]

where (h,k) is the center

The focus is

(h+p, k)

Equation of directrix is

x= h-p,

Here the center is (10,0)

Next, we factor out 4 in the original equation

[tex]4(9)[/tex]

So we have

[tex] {y}^{2} = 4(9)(x - 10)[/tex]

So our p=9,

So our focus is

(10+9,0) or (19,0)

The third option is the answer

The location of the light bulb is at point A. (10, 0).

Option A is the correct answer.

We have,

To find the location of the light bulb, we need to identify the coordinates (x, y) at which the light bulb is located.

In the given parabolic mirror equation, y² = 36(x - 10), the vertex form of the equation for a parabola with a vertical axis is (h, k), where h is the

x-coordinate of the vertex and k is the y-coordinate of the vertex.

Comparing the given equation with the standard form

(y - k)² = 4a(x - h),

we can see that h = 10 and k = 0.

So, the location of the light bulb is at the point (10, 0).

Therefore,

The location of the light bulb is at point A. (10, 0).

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A searchlight uses a parabolic mirror to cast a beam of light in parallel rays where the light bulb is located at the focus of the parabola in order to give the best illumination.

The mirror is modeled by y² = 36 (x-10), where the measurements are in cm.

What is the location of the light bulb?

A. (10, 0)

B. (14, 0)

C. (19, 0)

D. (36, 0)

The ways to collect data in a sample include:

It should be noted that a sample simply means a subset of a population that has the characteristics of the entire population.

In this case, the ways to collect data in a sample include the use of questionnaire, through observations, and through surveys.

l

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Answer:

10+1=11 but the 55 so it is now 11:55+15min so it makes it 12:10

Answer:

Hello! The answer to your question is 12:10.

Step-by-step explanation:

1 hour after 10:55 is equivalent to 11:55. Now, we have 15 minutes remaining. We can add in increments of 5 to get to 15 in terms of time:

11:55 + 5 minutes

= 12:00 + 5 minutes

= 12:05 + 5 minutes

= 12:10

5 + 5 + 5 = 15, so we reached 15.

The teacher collected the answer scripts at 12:10.

Answer:

[tex]a = (h)(2h - 5)[/tex]

Answer:

Area of the rectangle = 2h² - 5

Step-by-step explanation:

• base length l = 5 less than twice the height h

                        = 2h - 5

• height = h

Area of rectangle = base × height

                             ⇒ l × h

                             ⇒ (2h - 5)(h)

                              ⇒ 2h² - 5

Answer:

x = -6-2y

Step-by-step explanation:

This graph represents Option D = [tex]-3(x+2)^{2} + 4[/tex]

As the given figure is a parabola, it will have quadratic equation because the general equation of parabola corresponds  to [tex]x^{2} = 4ay[/tex] which has highest power of the variable equals to 2, thus its a quadratic equation.

The maximum value of a quadratic equation [tex]ax^{2} + b x +c[/tex]  is at points

[tex](-b/2a, c-\frac{b^{2} }{4a} )[/tex]

By looking at the graph we can observe the point at which the graph has achieved maxima is (-2,4).

∴ ([tex]-b/2a = -2[/tex] , [tex]c-\frac{b^{2} }{4a} = 4[/tex]

The coefficient of [tex]x^2[/tex] should be negative the graph opens downwards.

∴ Option B and C are eliminated.

On putting x = -2 only option D yields 4.

Thus this graph represents Option D = [tex]-3(x+2)^{2} + 4[/tex]

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The missing number from the triplet x, 48, 52, using the triplet 5, 12, 13, is found to be 5.

The Pythagoras Theorem states that in a right triangle, the square of the hypotenuse is always equal to the sum of the squares of the other two legs.

If the hypotenuse is taken as a, and the two legs are taken as b and c, then by the Pythagoras Theorem, we can write that:

a² = b² + c².

A Pythagoras triplet, is the combination of c, b, and a.

If all three sides of a right triangle are multiplied by the same quantity, we get another right triangle, where the sides of both make Pythagoras Triplet.

In the question, we are given a unique triplet: 5,12, 13, and are asked to find x in x, 48, 52.

We divide the second triplet by the first one, to get:

x/5, 48/12, 52/13,

or, x/5, 4, 4.

The ratio of the two triplets will always be a constant, thus

x/5 = 4,

or, x = 5*4 = 20.

Thus, the missing number from the triplet x, 48, 52, using the triplet 5, 12, 13, is found to be 5.

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The number of 1 foot cubic blocks that can be stuffed in the pyramid are; 10,583 cubes

The equation for the surface area of a square pyramid is;

Surface area = A + 1/2·p·s

where;

A = area of base

p = perimeter

s = slant height

The volume of the pyramid = (1/3) × A × h

Where:

A = Area of the base = 20 × 20 = 400 ft²

h = The height of the pyramid = √((slant height)² - ((base side length)/2)²)

h = √(80² - (20/2)²) = 30√7

The volume of the pyramid = 1/3*400*30·√7 = 4000·√7 ft³ = 10,583.005 ft³

If the cubes are flexible, then there are approximately 10,583 cubes

For rigid cubes we have;

Given that the height of the pyramid = 30·√7

The slope of the pyramid = (30/10)√7 = 3·√7

A increase in height of 1 foot gives a reduction in width of 2/(3√7)

The bottom can hold 400 cubes, Then next layer can hold;

19 × 19 = 361

If we continue to iterate, we will get a total of approximately 9915 rigid cubes in the pyramid.

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Question:

The perimeter of a square field is 336 yd. How long is each side?

Answer:

1,344 yd

Answer:

4.968

Step-by-step explanation:

bc yes

See below for the proof of the equation [tex]\frac{ax + by}{x + y} - \frac{ax - by}{x - y}= 2(a - b)[/tex]

The equation is given as:

[tex]\frac{ax + by}{x + y} - \frac{ax - by}{x - y}= 2(a - b)[/tex]

Take the LCM

[tex]\frac{(ax + by)(x - y) -(ax - by)(x + y)}{(x + y)(x - y)}= 2(a - b)[/tex]

Expand

[tex]\frac{ax^2 - axy + bxy - by^2 -ax^2 - axy + bxy + by^2}{(x + y)(x - y)}= 2(a - b)[/tex]

Evaluate the like terms

[tex]\frac{-2axy + 2bxy }{(x + y)(x - y)}= 2(a - b)[/tex]

Rewrite as:

[tex]\frac{-2axy + 2bxy }{x^2 - y^2}= 2(a - b)[/tex]

Factorize the numerator

[tex]\frac{2(a - b)(x^2 - y^2)}{x^2 - y^2}= 2(a - b)[/tex]

Divide

2(a - b)= 2(a - b)

Both sides are equal

Hence, the equation [tex]\frac{ax + by}{x + y} - \frac{ax - by}{x - y}= 2(a - b)[/tex] has been proved

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Answer:

17th term

Step-by-step explanation:

Substitute [tex]a_{n}[/tex] with the value 107

107=11+6(n-1)

96=6n-6

102=6n

n=17

17th term !!

Hope it helps!

Answer:

[tex]94.6cm^{2}[/tex]

Step-by-step explanation:

First we will find the area of Square PQRS.

Area of Square PQRS = Length x Width = 21 x 21

= [tex]441cm^{2}[/tex]

Next we will found the Area of Semicircles PS and QR.

Note: Area of Semicircle PS = Area of Semicircle QR

Area of Semicircle = [tex]\frac{1}{2} \pi r^{2}[/tex]

Total Area of Semicircles PS and QR combined = [tex]2(\frac{1}{2} \pi r^{2} )\\=\pi r^{2}[/tex]

We know that the diameter of PS = QR = 21 cm (due to the length of the square)

Radius = Half of Diameter = 0.5 x 21cm = 10.5cm

Total Area of Semicircles PS and QR = [tex]\pi (10.5)^{2} \\=110.25\pi cm^{2}[/tex]

Finally,

Area of Shaded Region = Area of Rectangle PQRS - Total Area of Semicircles PS and QR

= [tex]441 - 110.25\pi\\= 94.6cm^{2} (1dp)[/tex]

In this case , you can choose the nearest answer as there might be some rounding differences.

Answer:

x₁ = 4

(x₂ = -3)

Step-by-step explanation:

√(x + 12) = x

x + 12 = x²

x² - x - 12 = 0

Δ = 49

x = (1 ±√Δ)/2

x₁ = 4

x₂ = -3

the right solution is 4 this can be seen by inserting the solutions found in the starting equation