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SpyIntel [72]
3 years ago
7

Use a net to find the surface area of the right triangular prism shown below:

Mathematics
1 answer:
Bad White [126]3 years ago
6 0
The answer is 468............
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Which pair of triangles can be proven congruent by SAS?<br> AA
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The second one. There is a side, an angle, and another side.
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Khan academy question. Please help!
Zepler [3.9K]
It’s B
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Which two equations would be most appropriately solved by using the zero product property? Select each correct answer.
LekaFEV [45]
The zero product property tells us that if the product of two or more factors is zero, then each one of these factors CAN be zero.

For more context let's look at the first equation in the problem that we can apply this to: (x-3)(x+4)=0

Through zero property we know that the factor (x-3) can be equal to zero as well as (x+4). This is because, even if only one of them is zero, the product will immediately be zero.

The zero product property is best applied to factorable quadratic equations in this case.

Another factorable equation would be 2x^{2}+6x=0 since we can factor out 2x and end up with 2x(x+3)=0. Now we'll end up with two factors, 2x and (x+3), which we can apply the zero product property to.

The rest of the options are not factorable thus the zero product property won't apply to them.
3 0
3 years ago
Read 2 more answers
The equation a=1/2(b^1+b^2)h can be determined the area, a, of a trapezoid with height, h, and base lengths, b^1 and b^2 Which a
Evgesh-ka [11]

The complete question is as follows.

The equation a = \frac{1}{2}(b_1 + b_2 )h can be used to determine the area , <em>a</em>, of a trapezoid with height , h, and base lengths, b_1 and b_2. Which are equivalent equations?

(a) \frac{2a}{h} - b_2 = b_1

(b) \frac{a}{2h} - b_2 = b_1

(c) \frac{2a - b_2}{h} = b_1

(d) \frac{2a}{b_1 + b_2} = h

(e) \frac{a}{2(b_1 + b_2)} = h

Answer: (a) \frac{2a}{h} - b_2 = b_1; (d) \frac{2a}{b_1 + b_2} = h;

Step-by-step explanation: To determine b_1:

a = \frac{1}{2}(b_1 + b_2 )h

2a = (b_1 + b_2)h

\frac{2a}{h} = b_1 + b_2

\frac{2a}{h} - b_2 = b_1

To determine h:

a = \frac{1}{2}(b_1 + b_2 )h

2a = (b_1 + b_2)h

\frac{2a}{(b_1 + b_2)} = h

To determine b_2

a = \frac{1}{2}(b_1 + b_2 )h

2a = (b_1 + b_2)h

\frac{2a}{h} = (b_1 + b_2)

\frac{2a}{h} - b_1 = b_2

Checking the alternatives, you have that \frac{2a}{h} - b_2 = b_1 and \frac{2a}{(b_1 + b_2)} = h, so alternatives <u>A</u> and <u>D</u> are correct.

4 0
3 years ago
Help please explain in detail
yaroslaw [1]

Answer:

1. (n + 3)(5n + 8)

2. (x - 4)(7x - 4)

3. (k + 8)(7k + 1)

Step-by-step explanation:

1. We have to factorize 5n² + 23n + 24.

Now, 5n² + 23n + 24

= 5n² + 15n + 8n + 24

= 5n (n + 3) + 8 (n + 3)

=(n + 3)(5n + 8) (Answer)

2. We have to factorize 7x² - 32x + 16

Now,  7x² - 32x + 16

= 7x² - 28x - 4x + 16

= 7x (x - 4) - 4 (x - 4)

= (x - 4)(7x - 4) (Answer)

3. We have to factorize 7k² + 57k + 8

Now, 7k² + 57k + 8

= 7k² + 56k + k + 8

= 7k (k + 8) + 1 (k + 8)

= (k + 8)(7k + 1) (Answer)

4 0
3 years ago
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