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2 years ago

72÷(-18)-(-3)×4-[15-2{4+3(-6-5)}]Please help assignment is already past due!!!

Mathematics

1 answer:

Eduardwww [97]2 years ago

8 0

Answer: -65 is the right answer.

-65 is still the simplified answer btw

Hope this helps

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A cubic meter of water has a mass of 10 kilograms. How much water is this in grams?

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2 years ago

10 Pts + Brainliest for best answer.<br><br> Please answer quickly. Thanks!

Juliette [100K]

m∠AOC = m∠AOB + m∠BOC

We have m∠AOC = 19°, m∠AOB = 4x - 15 and m∠BOC = 2x + 10. Substitute:

19 = (4x - 15) + (2x + 10)

19 = (4x + 2x) + (-15 + 10)

19 = 6x - 5 <em>add 5 to both sides</em>

24 = 6x <em>divide both sides by 6</em>

4 = x

m∠BOC = 2x + 10 → m∠BOC = 2(4) + 10 = 8 + 10 = 18

m∠AOB = 4x - 15 → m∠AOB = 4(4) - 15 = 16 - 15 = 1

<h3>Answer: c. m∠BOC = 18°, m∠AOB = 1°</h3>

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2 years ago

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I need help!

blagie [28]

Here is the solution...

1) $\sqrt[3]{135} \\$

Solution

We can rewrite $\sqrt[3]{135} = \sqrt[3]{27} *\sqrt[3]{5}$

Here, $\sqrt[3]{27} = 3$

Now substitute the value, we get

$\sqrt[3]{135} = 3 \sqrt[3]{5}$

The answer is $3\sqrt[3]{5}$

2) $-5\sqrt[3]{40} Solution[tex]-5\sqrt[3]{40} = -5\sqrt[3]{8} *\sqrt[3]{5}$

Here $\sqrt[3]{8} = \sqrt[3]{2^{3} } = 2\\$

Therefore, we get $-5*2\sqrt[3]{5} = -10 \sqrt[3]{5}$

The answer is-10 \sqrt[3]{5}[/tex]

2) $2\sqrt[3]{5} * 4\sqrt[3]{8}$

Solution

$\sqrt[3]{8} = 2$

Now plug in the above in the given expression, we get

$2\sqrt[3]{5} * 4*2 = 16\sqrt[3]{5}$ {2*4*2 = 16]

The answer is $16\sqrt[3]{5}$

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3 years ago

72÷(-18)-(-3)×4-[15-2{4+3(-6-5)}]Please help assignment is already past due!!!​

72÷(-18)-(-3)×4-[15-2{4+3(-6-5)}]Please help assignment is already past due!!!