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zmey [24]
3 years ago
15

A photograph has a width of 9.5cm and an area of 104.5cm(squared). An enlargement is to be made that has a width of 19cm. What w

ill be the area of the enlargement?
Mathematics
1 answer:
SashulF [63]3 years ago
5 0
Answer: Area = 389.5cm^2

Explanation:

Area = 104.5cm^2
Width = 9.5cm

Find Length:

L = A/W
L = 104.5/9.5
L = 11cm

Find the enlargement:

let x be the enlargement:
9.5cm + x = 19cm
x = 19 - 9.5
x = 9.5cm

Thus, the enlargement is 9.5cm.

L (after enlarging) = 11 + 9.5 = 20.5cm

W (after enlarging) = 19cm

Find area after enlarging:

A = 20.5 x 19
A = 389.5cm^2

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Plz help ill give brainlyest
sashaice [31]

Answer:

your answer would be become 11 feet

5 0
3 years ago
PLEASE HELP GEOMETRY QUESTION: abcd and efgh are squares. if jh = 4 cm and jc = 9 cm, then what is the area of the shaded region
Nata [24]
Hello!

First you have to find the area of the bigger square

When it goes from the center to a corner you multiply the value by 2 to find one side

9 * 2 =18

Now you find the area

18 * 18 = 324

Now we find the area of the smaller region

4 * 2 = 8

Find the area

8 * 8 = 64

Now we subtract these

324 - 64 = 260

The answer is 260

Hope this helps!
3 0
4 years ago
Read 2 more answers
A poster is 8in by 6in what are the dimensions if the poster is enlarged by a factor of 7/2​
Bess [88]

Answer:

21\ in by 63\ in

Step-by-step explanation:

we know that

To find the dimensions of the enlarged poster, multiply the original dimensions of the poster by the scale factor

so

18(\frac{7}{2})=63\ in

6(\frac{7}{2})=21\ in

therefore

The dimensions of the enlarged poster are  21\ in by 63\ in

8 0
3 years ago
According to the Rational Root Theorem, which function has the same set of potential rational roots as the function g(x) = 3x5 –
SVETLANKA909090 [29]

We have to identify the function which has the same set of potential rational roots as the function g(x)= 3x^5-2x^4+9x^3-x^2+12.

Firstly, we will find the rational roots of the given function.

Let 'p' be the factors of 12

So, p= \pm 1, \pm 2, \pm 3, \pm 4, \pm 6

Let 'q' be the factors of 3

So, q=\pm 1, \pm 3

So, the rational roots are given by \frac{p}{q} which are as:

\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{4}{3}.

Consider the first function given in part A.

f(x) = 3x^5-2x^4-9x^3+x^2-12

Here also, Let 'p' be the factors of 12

So, p= \pm 1, \pm 2, \pm 3, \pm 4, \pm 6

Let 'q' be the factors of 3

So, q=\pm 1, \pm 3

So, the rational roots are given by \frac{p}{q} which are as:

\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{4}{3}.

Therefore, this equation has same rational roots of the given function.

Option A is the correct answer.

4 0
3 years ago
Read 2 more answers
Solve the following equation for x: 5x + 3y = 15.
zepelin [54]

Answer:

B) x = (-3/5)y + 3

Step-by-step explanation:

Isolate the variable, x. Note the equal sign, what you do to one side, you do to the other. Do the opposite of PEMDAS.

first, subtract 3y from btoh sides:

5x + 3y (-3y) = 15 (-3y)

5x = -3y + 15

Next, divide 5 from both sides:

(5x)/5 = (-3y +15)/5

x = (-3/5)y + 3

B) x = (-3/5)y + 3 is your answer.

~

8 0
3 years ago
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