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3 years ago

PLZ HELPPPPP!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics

1 answer:

stepan [7]3 years ago

5 0

1. One face is 5.1*2.5=12.75 un^2, one is 8.5*2.5=21.25, one is 6.8*2.5=17, and two are 1/2*6.8*5.1, so both are 34.68. Total is 85.68, rounded to 85.7 cm^2. (C)

2. The area of the circles is 2(π*2^2), which is about 25.1. The rectangle's area can be found as 2π*2*3, which is about 37.7. The total is rounded to 62.8 m^2 (D)

3. Rectangular faces: 3.6*1.8=6.48, 8.5*1.8=15.3, 7.7*1.8=13.86

Triangles: 2(1/2(3.6*7.7))=27.72

Total: 63.36, round to 63.4 km^2 (B)

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Urgent! Will award brainliest... Factor the following polynomial: <img src="https://tex.z-dn.net/?f=3a%5E%7B2%7D-21a%2B3

umka2103 [35]

Answer:

$3 (a-2) (a-5)$

Step-by-step explanation:

The first step is to find the GCF. Here, it's 3.

$3(a^2-7a+10)$

Then, you factor the polynomial in the parenthesis.

To find the factors, you will need to find 2 numbers that add to -7, and multiply to 10. -2 and -5 add to -7 and multiply to 10. Now, replace -7a with the factors.

$a^2-2a-5a+10$

This of this polynomial as 2 problems.

$a^2-2a$ $-5a+10$

Then, factor again.

$a^2-2a$ $-5a+10$

$a(a-2)$ $-5(a-5)$

Then, you keep the factors in parenthesis, and combine the numbers on the outside.

$(a-2)(a-5)(a-5)$

Since, there are 2 of the same factor, you only need one.

$(a-2)(a-5)$

BUT REMEMBER!! In the very beginning, we had a 3 that we took out, we STILL need to add that to the final answer. The final answer is:

$3 (a-2) (a-5)$

7 0

3 years ago

Over what interval will the immediate value theorem apply

koban [17]

Answer:

Any [a,b] that does NOT include the x-value 3 in it.

Either an [a,b] entirely to the left of 3, or

an [a,b] entirely to the right of 3

Step-by-step explanation:

The intermediate value theorem requires for the function for which the intermediate value is calculated, to be continuous in a closed interval [a,b]. Therefore, for the graph of the function shown in your problem, the intermediate value theorem will apply as long as the interval [a,b] does NOT contain "3", which is the x-value where the function shows a discontinuity.

Then any [a,b] entirely to the left of 3 (that is any [a,b] where b < 3; or on the other hand any [a,b] completely to the right of 3 (that is any [a,b} where a > 3, will be fine for the intermediate value theorem to apply.

6 0

2 years ago

Find g(a-2)+3g(2a) if g(x)=x²-5x+8

sertanlavr [38]

Answer:

13a² - 39a + 46

Step-by-step explanation:

To find g(a-2)+3g(2a), find each part using the function g(x)=x²-5x+8.

g(a-2) = (a-2)²-5(a-2)+8 = a² - 4a + 4-5a + 10+8 = a² - 9a + 22

3g(2a) = 3{(2a)²-5(2a)+8} = 3{ 4a² - 10a + 8} = 12a² - 30a + 24

Combine the values to find g(a-2)+3g(2a).

g(a-2)+3g(2a) = (a² - 9a + 22) + (12a² - 30a + 24) = 13a² - 39a + 46

4 0

3 years ago

3m - (7m+12) =2 (m-3)

viktelen [127]

Remove parentheses
3m - 7m+12 = 2 m-3

collect like terms
3m-7m-12 = 2m-6

move terms
-4m - 12 = 2m-6

collect the like terms and calculate
-4m-2m = -6+12

divide both sides by -6
-6m=6

m= -1

7 0

3 years ago

Simplify (-5) + (-4) + (7). 0-2 06 O 8 16

stich3 [128]

Answer:

- 2

Step-by-step explanation:

The question is - 5 + - 4 + 7

You do addition first.

-4 + 7 = 3

So now we have - 5 + 3

-5 + 3 = - 2

5 0

3 years ago