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2 years ago

Last year a bamboo plant was 17 feet tall. It grew 10 feet this year. How tall is it now?

Mathematics

2 answers:

Alekssandra [29.7K]2 years ago

8 0

It is 27 you add 10 and 27 which equals 27.

Vera_Pavlovna [14]2 years ago

4 0

Answer:

27 is the right answer

Step-by-step explanation:

10+17=27 :)

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Write an expression for the missing value in

Snowcat [4.5K]

C. a + 3

10 + 3 = 13
11 + 3 = 14
12 + 3 = 15
a + 3 = ?

5 0

3 years ago

Read 2 more answers

The number of eggs in the refrigerator e decreased by 9 equals 17.

alexandr402 [8]

Answer:

x-9=17

Step-by-step explanation:

x=eggs in the fridge

5 0

2 years ago

The value of is between 4.7 and 4.8. Which of the following is a more precise approximation of ?

Nataly_w [17]

Answer: D. between 4.79 and 4.8

Step-by-step explanation:

The square root of 23 (√23) is 4.79583152331. We can obviously eliminate A and B. C and D is left. C is wrong because if we approximate the square root of 23, which is 4.79583152331, we get 4.79. 4.79 > 4.78, so it can't be C. Hope this helps :)

4 0

3 years ago

Rewrite 9cos 4x in terms of cos x.

rosijanka [135]

$\bf \qquad \textit{Quad identities}\\\\
sin(4\theta )=
\begin{cases}
8sin(\theta )cos^3(\theta )-4sin(\theta )cos(\theta )\\
4sin(\theta )cos(\theta )-8sin^3(\theta )cos(\theta )
\end{cases}
\\\\\\
cos(4\theta)=8cos^4(\theta )-8cos^2(\theta )+1\\\\
-------------------------------\\\\
9cos(4x)\implies 9[8cos^4(x)-8cos^2(x)+1]
\\\\\\
72cos^4(x)-72cos^2(x)+9$

---------------------------------------------------------------------------

as far as the previous one on the 2tan(3x)

$\bf tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\qquad tan({{ \alpha}} + {{ \beta}}) = \cfrac{tan({{ \alpha}})+ tan({{ \beta}})}{1- tan({{ \alpha}})tan({{ \beta}})}\\\\
-------------------------------\\\\$

$\bf 2tan(3x)\implies 2tan(2x+x)\implies 2\left[ \cfrac{tan(2x)+tan(x)}{1-tan(2x)tan(x)}\right]
\\\\\\
2\left[ \cfrac{\frac{2tan(x)}{1-tan^2(x)}+tan(x)}{1-\frac{2tan(x)}{1-tan^2(x)}tan(x)}\right]\implies 2\left[ \cfrac{\frac{2tan(x)+tan(x)-tan^3(x)}{1-tan^2(x)}}{\frac{1-tan(x)-2tan^3(x)}{1-tan^2(x)}} \right]
\\\\\\$

$\bf 2\left[ \cfrac{2tan(x)+tan(x)-tan^3(x)}{1-tan^2(x)}\cdot \cfrac{1-tan^2(x)}{1-tan(x)-2tan^3(x)} \right]
\\\\\\
2\left[ \cfrac{3tan(x)-tan^3(x)}{1-tan^2(x)-2tan^3(x)} \right]\implies \cfrac{6tan(x)-2tan^3(x)}{1-tan^2(x)-2tan^3(x)}$

4 0

2 years ago

Jamal has at most $100 to spend on computer paper and printer ink for his computer. He buys 1 box of paper for $24.99. If ink ca

lions [1.4K]

Answer:

1 box of paper with 3 ink cartridges

Step-by-step explanation:

$100 total

$24.99+19.99(3)= $84.96 (4 ink cartridges would be over the allotted amount)

7 0

3 years ago