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3 years ago

Please help out explanation need it

Mathematics

1 answer:

prisoha [69]3 years ago

8 0

Answer:

A=2(wl+hl+hw)

A=2(6ft×7ft+6ft×7ft+6ft×6ft)

A=2(42ft²+42ft²+36ft²)

A=2(120ft²)

A=240ft²

Step-by-step explanation:

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A construction worker earns 2880 in 6 days. If he earns 8640 in a month, for how

V125BC [204]

Step-by-step explanation:

2880x3=8640

6x3=18

he worked 18 days.

6 0

3 years ago

Please answer i need it quick :(

Oksi-84 [34.3K]

Not sure but I think its 120
I think 120 times 2 and I got 240. Then I added the number of oak trees which was 120 and got 360

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3 years ago

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Which expression will determine the perimeter, in inches, of the rectangle? O 10 x 7 10 + 7 O (2 x 7) + (2 x 10) O 10 x 7 x 7 x

fredd [130]

Answer:

$(2 * 7) + (2 * 10)$

Step-by-step explanation:

The missing details in the question are:

$Length = 10$

$Width = 7$

Required

The perimeter

This is calculated as:

$Perimeter = 2 * (Length + Width)$

So, we have:

$Perimeter = 2 * (10 + 7)$

Open bracket

$Perimeter = 2 * 10 + 2 * 7$

Rewrite as:

$Perimeter = (2 * 7) + (2 * 10)$

3 0

2 years ago

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Choose the correct simplification of f^9 h^23/f^3 h^17

Tanzania [10]

Answer:

f6h40

Step-by-step explanation:

Step 1 :

h23

Simplify ———

Equation at the end of step 1 :

h23

((f9) • ———) • h17

Step 2 :

Multiplying exponential expressions :

2.1 h23 multiplied by h17 = h(23 + 17) = h40

Final result :

f6h40

4 0

3 years ago

To find the measure of ∠A in ∆ABC, use the___(Pythagorean Theorem, Law of Sines, Law of Cosines). To find the length of side HI

nadya68 [22]

Part 1) To find the measure of ∠A in ∆ABC, use

we know that

In the triangle ABC

Applying the law of sines

$\frac{a}{sin\ A}=\frac{b}{sin\ B}=\frac{c}{sin\ C}$

in this problem we have

$\frac{a}{sin\ A}=\frac{b}{sin\ theta}\\ \\a*sin\ theta=b*sin\ A\\ \\ sin\ A=\frac{a*sin\ theta}{b} \\ \\ A=arc\ sin (\frac{a*sin\ theta}{b})$

therefore

the answer Part 1) is

Law of Sines

Part 2) To find the length of side HI in ∆HIG, use

we know that

In the triangle HIG

Applying the law of cosines

$g^{2}=h^{2}+i^{2}-2*h*i*cos\ G$

In this problem we have

g=HI

G=angle Beta

substitute

$HI^{2}=h^{2}+i^{2}-2*h*i*cos\ Beta$

$HI=\sqrt{h^{2}+i^{2}-2*h*i*cos\ Beta}$

therefore

the answer Part 2) is

Law of Cosines

3 0

3 years ago

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