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S_A_V [24]
2 years ago
11

64 decreased by twice Craig’s savings

Mathematics
2 answers:
Ber [7]2 years ago
4 0
I’m just going to use x to represent Craig’s savings, since I don’t have any clue as to what that is.

64-2x
SSSSS [86.1K]2 years ago
3 0

71 - 2n I think I hope I'm right

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See image<br> thanks so much!!!<br> it’s about dilations
pshichka [43]
Use x,y coordinate
3 0
3 years ago
It costs $57.19 to fill up a 19 gallon gas tank and $54.18 to fill up an
Anton [14]

Answer:

y=3.01x

Step-by-step explanation:

We can see that since 19 gallons cost $57.19 and 18 gallons cost $53.18, 1 gallon (19-18) must cost $3.01.

Therefore, y=3.01x

You can check this with algebra

8 0
2 years ago
Evaluate the expression without using the calculator 8^1/3
allsm [11]

Answer:

= 2.67

Step-by-step explanation:

Given: 8^1/3

We have to find the value of the given fraction which can be done by dividing the denominator value with the numerator value

                                  \frac{8^1}{3} \\

 The power of 8 is '1', so, the value of numerator is 8.

                                    \frac{8}{3}

The value of the fraction is calculated by dividing 3 by 8.

                         = 2.67

8 0
3 years ago
Describe the steps you would use to solve the equation 4x + 7 = 15
choli [55]

i would subtract 7 and divide by 4;

4x + 7 = 15

4x = 8

x = 2

8 0
2 years ago
Read 2 more answers
In Problems 23–30, use the given zero to find the remaining zeros of each function
Talja [164]

Answer:

x =  2i, x = -2i and x = 4 are the roots of given polynomial.

Step-by-step explanation:

We are given the following expression in the question:

f(x) = x^3 - 4x^2+ 4x - 16

One of the zeroes of the above polynomial is 2i, that is :

f(x) = x^3 - 4x^2+ 4x - 16\\f(2i) = (2i)^3 - 4(2i)^2+ 4(2i) - 16\\= -8i+ 16+8i-16 = 0

Thus, we can write

(x-2i)\text{ is a factor of polynomial }x^3 - 4x^2 + 4x - 16

Now, we check if -2i is a root of the given polynomial:

f(x) = x^3 - 4x^2+ 4x - 16\\f(-2i) = (-2i)^3 - 4(-2i)^2+ 4(-2i) - 16\\= 8i+ 16-8i-16 = 0

Thus, we can write

(x+2i)\text{ is a factor of polynomial }x^3 - 4x^2 + 4x - 16

Therefore,

(x-2i)(x+2i)\text{ is a factor of polynomial }x^3 - 4x^2 + 4x - 16\\(x^2 + 4)\text{ is a factor of polynomial }x^3 - 4x^2 + 4x - 16

Dividing the given polynomial:

\displaystyle\frac{x^3 - 4x^2 + 4x - 16}{x^2+4} = x -4

Thus,

(x-4)\text{ is a factor of polynomial }x^3 - 4x^2 + 4x - 16

X = 4 is a root of the given polynomial.

f(x) = x^3 - 4x^2+ 4x - 16\\f(4) = (4)^3 - 4(4)^2+ 4(4) - 16\\= 64-64+16-16 = 0

Thus, 2i, -2i and 4 are the roots of given polynomial.

4 0
3 years ago
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