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just olya [345]
3 years ago
11

Please help thank u

Mathematics
1 answer:
puteri [66]3 years ago
6 0

Answer:

2 {x}^{2}  + 13x + 15 \\ 2 {x}^{2}  + 10x + 3x + 15 \\ 2x(x + 5) + 3(x + 5) \\ (2x + 3)(x + 5)

5 {x}^{2} +  21x + 4 \\ 5 {x}^{2}  + 20x + x + 4 \\ 5x(x + 4) + 1(x + 4) \\ (5x + 1)(x + 4) \\

7 {x}^{2}  - 59x + 24 \\  7 {x}^{2}  - 56x - 3x + 24 \\ 7x(x - 8) - 3(x - 8) \\ (7x - 3)(x - 8)

2 {x}^{2}  + 11x - 13 \\ 2 {x}^{2}   - 2x + 13x - 13 \\ 2x(x - 1) + 13(x - 1) \\ (2x + 13)(x - 1)

8 {x}^{2}  + 29x - 12 \\ 8 {x}^{2}  + 32x - 3x - 12 \\ 8x(x + 4) - 3(x + 4) \\ (8x - 3)(x + 4) \\  \\

12 {x}^{2}  - 7x - 12 \\ 12 {x}^{2}   + 9x - 16x- 12 \\3x(4x + 3) - 4(4x + 3) \\ (3x - 4)(4x + 3)

- 12 {x}^{2}  - 35x - 18 \\  - 12 {x}^{2}  - 8x - 27x- 18 \\  - 4x(3x - 2) - 9(3x - 2) \\ ( - 4x - 9)(3x - 2)

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For my math it says someone ordered a poster with a height of 2 feet and width of 0.7 feet what is the area of the poster
Phantasy [73]
STEP-BY-STEP SOLUTION:

Let's solve this problem step-by-step.

As posters are generally rectangular in shape, we will establish the formula for the area of a rectangle which we will be using as displayed below:

Area = height × width

Now let's find the area of the postwr using the above formula.

Height = 2 feet

Width = 0.7 feet

Area = height × width

Area = 2 × 0.7

Area = 1.4 square feet

ANSWER:

Therefore, the answer is:

The area of the poster is 1.4 square feet.

Please mark as brainliest if you found this helpful! <3
Thank you : )
6 0
3 years ago
John shot a toy rocket into the air. Consider the graph below which shows the relationship between the height of the rocket from
EleoNora [17]
<h2>The height of the rocket increases for some time and then decreases for some time.</h2>

The height from the ground increases from 4 to 26, then decreases from 26 to 0.

Why the others are wrong.

A. The height of the rocket changes at a constant rate for the entire time.

The graph is a curve. This means the rate is not constant. If it were constant, the graph would be linear - a straight line.

C. The height of the rocket remains constant for some time.

The graph is a curve. This means the rate is not constant. If it were constant, the graph would be linear - a straight line.

D. The height of the rocket decreases for some time and then increases for some time.

This implies the graph decreases first then increases. However, the rocket will increase, then decrease.

5 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20%5Clarge%7B%5Cbold%20%5Cred%7B%20%5Csum%20%5Climits_%7B8%7D%5E%7B4%7D%20%7Bx%7D%5E%7B2%7D%2
kiruha [24]

Answer:

No solution

Step-by-step explanation:

We have

$\sum_{x=8}^{4}x^2 + 9\left(\dfrac{\cos^2(\infty)}{\sin^2(\infty)} \right)$

For the sum it is not correct to assume

$\sum_{x=8}^{4}x^2= 8^2 + 7^2+6^2+5^2+4^2 = 64+49+36+25+16 = 190$

Note that for

$\sum_{x=a}^b f(x)$

it is assumed a\leq x \leq b and in your case \nexists x\in\mathbb{Z}: a\leq x\leq b for a>b

In fact, considering a set S we have

$\sum_{x=a}^b (S \cup \varnothing) = \sum_{x=a}^b S + \sum_{x=a}^b \varnothing $ that satisfy S = S \cup \varnothing

This means that, by definition \sum_{x=a}^b \varnothing = 0

Therefore,

$\sum_{x=8}^{4}x^2 = 0$

because the sum is empty.

For

9\left(\dfrac{\cos^2(\infty)}{\sin^2(\infty)} \right)

we have other problems. Actually, this case is really bad.

Note that \cos^2(\infty) has no value. In fact, if we consider for the case

$\lim_{x \to \infty} \cos^2(x)$, the cosine function oscillates between [-1, 1] , and therefore it is undefined. Thus, we cannot evaluate

9\left(\dfrac{\cos^2(\infty)}{\sin^2(\infty)} \right)

and then

$\sum_{x=8}^{4}x^2 + 9\left(\dfrac{\cos^2(\infty)}{\sin^2(\infty)} \right)$

has no solution

7 0
3 years ago
Find the area of the shaded region. explain your answer.​
andrezito [222]

Answer:

20cm

Step-by-step explanation:

A=hb/2

h=5cm

b=8cm

A=(5*8)/2

A=40/2

A=20

8 0
3 years ago
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Fiesta28 [93]

Answer:

x=4

Step-by-step explanation:

subtract 8 from both sides

then you get 2x=8

then divide by 2

then you get x=4

6 0
3 years ago
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