Find the equation for the circle with center (8,−8) and radius of length 2

4 3/4 = 19/4

133/(19/4) = 133(4/19) = 1197/19 = 63 mph

Answer:

The answer is 4

Step-by-step explanation:

(-2.4)(10) / -3(2)

-24/-6

=4

Answer:

The answer will be 4! Hope this helps

Answer:

-1.5

Step-by-step explanation:

Answer:

1.2

Step-by-step explanation:

First, you cancel out [tex]\frac{6}{20}[/tex] to become [tex]\frac{3}{10}[/tex]

Then that becomes [tex]-\frac{3}{10}+1.5[/tex]

Now, you convert the element to a decimal form to be [tex]-0.3+1.5[/tex]

Finally, you subtract them (because a negative plus a positive is subtraction) to get the final answer of 1.2

Hope this helps!

Answer: 8.5

Step-by-step explanation: (42.50 - 17) ÷ (5 - 2)

= 25.50 ÷ 3

= 8.5

Thank me later

The strategy that won't help her to calculate 9 - 7 is (a) Make 10

The given parameters are:

[tex]Yellow = 7[/tex]

[tex]Red = 9[/tex]

To determine 9 - 7,

However, she cannot make 10.

This is so, because 9 and 7 are not up to 10

Hence, the strategy that won't help her to calculate 9 - 7 is (a)

Read more about subtraction at:

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[tex]y = x^{\tfrac 1{\ln x}}\\\\\implies \ln y = \ln \left( x^{\tfrac 1{\ln x}} \right)\\\\\implies \ln y = \dfrac{1}{\ln x} \cdot \ln x\\\\\implies \ln y = 1\\\\\implies y =e\\\\\\\text{Now,}\\\\\ \lim \limits_{x \to \infty} x^{\tfrac 1{\ln x}}=\lim \limits_{x \to \infty} y = \lim \limits_{x \to \infty} e = e[/tex]

Answer:

2

Ok lets start off with your 3 numbers

First do 10 / 5 = 2

Then do 12 / 6 = 2

After that do 20 / 10 = 2

now you see you that its all equivalent to 2 that's your answer!

Answer:

d.

Step-by-step explanation:

[tex]x = 23[/tex]

Step-by-step explanation:

First, we need to write the equations in their standard form, i.e., all the variables and their coefficients are to be placed on the left hand side and the plain numbers on the right hand side. The first equation is already in its standard form:

[tex]2x + 5y = 35\;\;\;\;\;(1)[/tex]

but we need to rewrite the 2nd one. So the standard form of the 2nd equation can be written as

[tex]x + 15y = -10\;\;\;\;\;(2)[/tex]

Now that the equations are in their standard forms, the next step that we need to do is to eliminate one of the variables, namely the y. To do that, we need to multiply either one of the equations so that when we add them together, the y-variable gets eliminated. Note if we multiply Eqn(1) by -3, we'll get -15y on one equation and a +15y on the other equation so that when they are added together, they cancel out. So if we do that, we get

[tex]-6x - 15y = -105\;\;\;\;\;(3)[/tex]

Adding Eqn(3) to Eqn(2), we get

[tex]-5x = -115[/tex]

or

[tex]x = \dfrac{-115}{-5} = 23[/tex]

[tex]\dfrac 25 (d-10) - \dfrac 23(d+6)\\\\\\=\dfrac{6}{15} (d-10) - \dfrac{10}{15} (d+6)\\\\\\=\dfrac{6(d-10) -10(d+6)}{15}\\\\\\=\dfrac{6d -60-10d-60}{15}\\\\\\=\dfrac{-4d-120}{15}\\\\\\=-\dfrac{4d}{15} - 8[/tex]

Answer:

<

Step-by-step explanation:

Good Question!

8/20=4/10=2/5=0.4

7/13= lcm of 5 and 13 is 75.

26/75 [<] 35/75

:)

Answer: < less than

Step-by-step explanation:

1. Find LCM of 20 and 13.

20: 20,40,60,80,100,120,140,160,180,200,220,240,260,280

13: 13, 26, 39, 52, 65, 78, 91, 104, 107, 110 ...... 234, 247, 260

The LCM is 260.

2. 20*13 = 260 13*20=260

therefore we need to multiply the numerator by those 13 and 20.

8*13 = 104 7*20 = 140

3. compare Fractions:

(104/260) < (140/260)

[tex]\begin{cases} x = -4\implies &x+4=0\\ x=-2\implies &x+2=0\\ x=1\implies &x-1=0 \end{cases}\qquad \implies (x+4)(x+2)(x-1)=\stackrel{y}{0}[/tex]

now, that's the equation or polynomial in factored form, hmmm we also know that it has a y-intercept of -11, namely, when x = 0  y = -11, well let's plug in a factor to it, that will reflect those values, namely say hmmm factor "a", so

[tex](x+4)(x+2)(x-1)=y\qquad \stackrel{\textit{adding "a" factor for vertical shift}}{a(x+4)(x+2)(x-1)}=y \\\\\\ \stackrel{\textit{we know that when x = 0, y = -11}}{a(0+4)(0+2)(0-1)=-11}\implies -8a=-11\implies a=\cfrac{-11}{-8}\implies a = \cfrac{11}{8} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\mathbb{FOIL}}{\cfrac{11}{8}(x^2+6x+8)}(x-1)=y\implies \cfrac{11}{8}(x^3+6x^2+8x-x^2-6x-8)=y \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \cfrac{11}{8}(x^3+5x^2+2x-8)=y~\hfill[/tex]

Answer:

$1650 is the answer hope it helps

Step-by-step explanation:

although there is no graph visible here, I suspect this is the same question as I just answered twice already.

the slope is -6.

D is the right answer.

Answer:

What's the question lol

Answer:

Yes.

Step-by-step explanation:

Associative property of multiplication is:

(A×B)×C = A×(B×C).

It says that "The way of grouping of terms doesn't change the product."

So, given example (2×3)×4=2×(3×4) is an example of associative property of multiplication.

[tex]\rule[225]{225}{2}[/tex]

Hope this helped!

Answer:

[tex]x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}}[/tex] and [tex]y=-1,\frac{1}{2}[/tex]

The ordered pair solutions are [tex](-\sqrt{\frac{7}{4}},0.5)[/tex], [tex](\sqrt{\frac{7}{4}},0.5)[/tex], [tex](-1,-1)[/tex], and [tex](1,-1)[/tex].

Step-by-step explanation:

I'm assuming the system is [tex]\left \{ {x^2+y^2=2} \atop {y=2x^2-3}} \right.[/tex]:

[tex]x^2+y^2=2[/tex]

[tex]x^2+(2x^2-3)^2=2[/tex]

[tex]x^2+(4x^4-12x^2+9)=2[/tex]

[tex]x^2+4x^4-12x^2+9=2[/tex]

[tex]4x^4-11x^2+9=2[/tex]

[tex]4x^4-11x^2+7=0[/tex]

[tex]x^4-11x^2+28=0[/tex]

[tex](x^2-7)(x^2-4)=0[/tex]

[tex](4x^2-7)(x^2-1)=0[/tex]

[tex]4x^2-7=0[/tex]

[tex]4x^2=7[/tex]

[tex]x^2=\frac{7}{4}[/tex]

[tex]x=\pm\sqrt{\frac{7}{4}}[/tex]

[tex]x^2-1=0[/tex]

[tex]x^2=1[/tex]

[tex]x=\pm1[/tex]

[tex]y=2x^2-3[/tex]

[tex]y=2(\pm\sqrt{\frac{7}{4}})^2-3[/tex]

[tex]y=2({\frac{7}{4}})-3[/tex]

[tex]y=\frac{7}{2}-3[/tex]

[tex]y=\frac{1}{2}[/tex]

[tex]y=2x^2-3[/tex]

[tex]y=2(\pm1)^2-3[/tex]

[tex]y=2(1)-3[/tex]

[tex]y=2-3[/tex]

[tex]y=-1[/tex]

Therefore, [tex]x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}}[/tex] and [tex]y=-1,\frac{1}{2}[/tex]

The ordered pair solutions are [tex](-\sqrt{\frac{7}{4}},0.5)[/tex], [tex](\sqrt{\frac{7}{4}},0.5)[/tex], [tex](-1,-1)[/tex], and [tex](1,-1)[/tex].

Answer:

  quotient term: 6

  dividend term: 102

Step-by-step explanation:

[tex]\dfrac{782}{17}=\dfrac{680+102}{17}=\dfrac{680}{17}+\dfrac{\boxed{102}}{17}=40 +\boxed{6} = 46[/tex]

Check the picture below.

The value of f(x) at x=4 is 2.

A function is an expression, rule, or law in mathematics that describes a relationship between one variable (the independent variable) and another variable (the dependent variable). Functions are common in mathematics and are required for the formulation of physical relationships in the sciences.

We have to find the value of function for x=4.

This means that from the figure we have to find the value of y when x= 4.

Now, looking into the figure we have

f(4)= 2

f(0)= -2

f(-1)= -1

f(3)= 4

So, the required value is 2.

Learn more about Function here:

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#SPJ7

Answer:

16x7 = 112 (16 ounces, 7 hours) 112 ounces to pints = 7 pints of water

Step-by-step explanation:

Answer:

The sum of 8 and 1 is 9.

lol

Step-by-step explanation:

Answer:

I think this is what is wanted

Step-by-step explanation:

Answer:

[tex]{ \rm{V = \frac{lwh}{3} }} \\ \\ { \rm{3V = lwh}} \\ \\ { \boxed{ \rm{ \: w = \frac{3V}{lh } \: }}}[/tex]

Formula:

We know that

nth term of the AP is Tn = a+(n-1)d

Where, a = First term

d = Common difference

n = number of terms

Now,

Given that

pth term of the A.P. = Tp = q

⇛a + (p-1)d = q

⇛(p-1)d = q - a

⇛d = (q-a)/(p-1) →→→→(i)

and

qth term of the A.P. = Tq = p

⇛a + (q-1) d = p

⇛ (q-1)d = p-a

⇛ d = (p-a)/(q-1) →→→→(ii)

From Eqn(i) and Eqn(ii)

⇛(q-a)/(p-1) = (p-a)/(q-1)

On applying cross multiplication then

⇛(q-a)×(q-1) = (p-a)×(p-1)

⇛q²-q-aq+a = p²-p-ap+a

⇛q²-q-aq = p²-p-ap

⇛ap-aq = p²-p+q-q²

⇛a(p-q) = (p²-q²)-(p-q)

⇛a(p-q) = (p+q)(p-q)-(p-q)

⇛ a(p-q) = (p-q){(p+q)-1}

On cancelling (p-q) both sides then

⇛a = (p+q-1) →→→→Eqn(iii)

On Substituting the value of a in Eqn(ii) then

⇛ d = [p-(p+q-1)]/(q-1)

⇛d = (p-p-q+1)/(q-1)

⇛d = (-q+1)/(q-1)

⇛d = -(q-1)/(q-1)

⇛d = -1

We have,

a = p+q-1 and d = -1

Now, rth term of the AP

⇛ Tr = a + (r-1)d

⇛Tr = (p+q-1)+(r-1)(-1)

⇛ Tr = p+q-1+(-r+1)

⇛ Tr = p+q-1-r+1

⇛ Tr = p+q-r

Therefore, Tr = p+q-r

Answer: rth term of the given A.P. is p+q-r

Answer:

rth=q+p+qxp

Step-by-step explanation:

rth= 3 of the T and Tq added together and joined.