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3 years ago

Cody was 165cm tall on the first day of school this year, which was 10% taller than he was on the first day of school last year.

Mathematics

1 answer:

Solnce55 [7]3 years ago

7 0

On the first day of school last year, Cody was 148.5 cm tall.

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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

3 years ago

Which is true about the product of 3/8 and 7/2. 1.The product is less than 3/8. 2.The product is greater than 1/8 and less that

GalinKa [24]

Answer:

4.The product is greater that 3/8 and less than 7/2.

Step-by-step explanation:

The product of 3/8 and 7/2 is:

$\frac{3}{8} * \frac{7}{2} \\\\= \frac{21}{16} = 1\frac{5}{16}$

This number is greater than 3/8.

It is greater than 1/8 but not less than 1/2.

This number is less than 7/2.

This number is greater than 3/8 and it is less than 7/2.

Hence, the correct options is 4.

7 0

2 years ago

Find the slope and reduce of p=(3/4,1 1/4) q=(-1/2,-1)

sergij07 [2.7K]

$\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
p&({{ \frac{3}{4}}}\quad ,&{{ \frac{1}{4}}})\quad 
% (c,d)
q&({{ -\frac{1}{2}}}\quad ,&{{ -1}})
\end{array}$

$\bf slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{-1-\frac{1}{4}}{-\frac{1}{2}-\frac{3}{4}}\quad \cfrac{\leftarrow LCD=4}{\leftarrow LCD=4}
\\\\\\
\cfrac{\frac{-4-1}{4}}{\frac{-2\cdot 1-3}{4}}\implies \cfrac{\frac{-5}{4}}{\frac{-5}{4}}\implies \cfrac{-5}{4}\cdot \cfrac{4}{-5}\implies 1$

7 0

3 years ago

Simplify: 2cd3 (c- d+5)

sasho [114]

Answer:

2c2d3 − 2cd4 + 10cd3

hope it helps

6 0

3 years ago

Sofia is working two summer jobs, making $12 per hour babysitting and making $8 per hour clearing tables. In a given week, she c

Mamont248 [21]

The possible values for the number of whole hours clearing tables that she must work to meet her requirements is 2, 3 hours

<em><u>Solution:</u></em>

Amount earned in babysitting = $ 12 per hour

Amount earned in clearing tables = $ 8 per hour

In a given week, she can work a maximum of 17 total hours and must earn a minimum of $180

Sofia worked 14 hours babysitting

Therefore,

Amount earned at babysitting = 14 x 12 = 168

Thus, Sofia earned $ 168 at babysitting

Sofia must earn a minimum of $ 180

Remaining amount to be earned = 180 - 168 = 12

Thus, Sofia must earn $ 12 from clearing tables

Amount earned in clearing tables = $ 8 per hour

So, she must work for atleast 1.5 hours to get $ 12 from clearing tables

She can work a maximum of 17 total hours and Sofia worked 14 hours babysitting

Remaining is 17 - 14 = 3 hours

Thus possible values for the number of whole hours clearing tables that she must work to meet her requirements is 2 hours or 3 hours

3 0

3 years ago