Help me help me help me

morpeh [17]

I need a photo but ill give you some information this will be with exponents.

Write down the problem. Here it is:

2^2(x+3) + 9 - 5 = 32

Resolve the exponent. Remember the order of operations: PEMDAS, which stands for Parentheses, Exponents, Multiplication/Division, and Addition/Subtraction. You can't resolve the parentheses first because x is in the parentheses, so you should start with the exponent, 2^2.2^2 = 4

4(x+3) + 9 - 5 = 32

Do the multiplication. Just distribute the 4 into (x +3). Here's how:

4x + 12 + 9 - 5 = 32

Do the addition and subtraction. Just add or subtract the remaining numbers. Here's how:

4x+21-5 = 32

4x+16 = 32

4x + 16 - 16 = 32 - 16

4x = 16

Isolate the variable. To do this, just divide both sides of the equation by 4 to find x. 4x/4 = x and 16/4 = 4, so x = 4.

4x/4 = 16/4

x = 4

Check your work. Just plug x = 4 back into the original equation to make sure that it checks out. Here's how:

2^2(x+3)+ 9 - 5 = 32

2^2(4+3)+ 9 - 5 = 32

2^2(7) + 9 - 5 = 32

4(7) + 9 - 5 = 32

28 + 9 - 5 = 32

37 - 5 = 32

32 = 32

Credit to wikiHow.com

7 0

3 years ago

25 POINTS AVAILABLE

myrzilka [38]

Answer:

$\large\boxed{1.\ (-3, 0),\ r = 3}\\\boxed{2.\ (x+4)^2+(y-3)^2=36}\\\boxed{3.\ (x-2)^2+(y-1)^2=(\sqrt{34})^2}$

Step-by-step explanation:

The equation of a circle in standard form:

$(x-h)^2+(y-k)^2=r^2$

(h, k) - center

r - radius

1. We have the equation:

$(x+3)^2+y^2=9\\\\(x-(-3))^2+(y-0)^2=3^2$

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2. We have the center (-4, 3) and the radius r = 6. Substitute:

$(x-(-4))^2+(y-3)^2=6^2\\\\(x+4)^2+(y-3)^2=36$

3. We have the endpoints of the diameter: (-1, 6) and (5, -4).

Midpoint of diameter is a center of a circle.

The formula of a midpoint:

$\left(\dfrac{x_1+x_2}{2};\ \dfrac{y_1+y_2}{2}\right)$

Substitute:

$h=\dfrac{-1+5}{2}=\dfrac{4}{2}=2\\\\k=\dfrac{6+(-4)}{2}=\dfrac{2}{2}=1$

The center is in (2, 1).

The radius length is equal to the distance between the center of the circle and the endpoint of the diameter.

The formula of a distance between two points:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Substitute the coordinates of the points (2, 1) and (5, -4):

$r=\sqrt{(5-2)^2+(-4-1)^2}=\sqrt{3^2+(-5)^2}=\sqrt{9+25}=\sqrt{34}$

Finally we have:

$(x-2)^2+(y-1)^2=(\sqrt{34})^2$

7 0

3 years ago

a rule for an example of a nonlinear function that fits the following description. when d is 4, r is 9, and r is a function of d

Anika [276]

Some possibilities:
r = (9√d)/2, so f(d) = (9√d)/2
r = 9d^2/16, so f(d) = 9d^2/16
r = 9(2^d)/16, so f(d) = 9(2^d)/16

These are examples of non-linear functions of d which fit the description. I think that is what you are asking, but comment if I'm wrong?

3 0

3 years ago

Liam has 2 quarts of apple juice he wants to pour the juice into 1/5 quart servings how many servings can Liam pour

Delicious77 [7]

The answer would be 10.
You would divide 2 divided by 1/5.

4 0

3 years ago

Read 2 more answers

Find the first, fourth, and tenth terms of the arithmetic sequence described by the given rule.

lina2011 [118]

Answer is in the photo. I can only upload it to a file hosting service. link below!

tinyurl.com/wpazsebu