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Dahasolnce [82]

3 years ago

5

A box contains 5 plain pencils and 5 pens. A second box contains 5 color pencils and 7 crayons. One item from each box is chosen

at random

2 answers:

chubhunter [2.5K]3 years ago

6 0

Is there more because with out it we can’t understand what it means

bazaltina [42]3 years ago

6 0

Can you give more information on the question

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A rectangular prism has a length of 4.2 cm, a width of 5.8 cm, and a height of 9.6 cm. A similar prism has a length of 14.7 cm,

solong [7]

They are each multiplied by a factor of 3.5
You can calculate this by taking the length of the similar prism and divide it with the first prism 14.7/4.2 = 3.5
Do this with the width and height to be sure
20.3/5.8 = 3.5
33.6/9.6 = 3.5

5 0

3 years ago

Read 2 more answers

I need to know how to solve ASAP!! Thanks!

Viktor [21]

Answer:

x=24x

Step-by-step explanation:

Brainly

8 0

2 years ago

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Expressions are equivalent???

Semenov [28]

Answer:

only the first option is equivalent

Step-by-step explanation:

4 0

3 years ago

Which of the following expressions can be factored to 2?

Ber [7]

Answer:

Correct answer is A.

Step-by-step explanation:

2x∧3 + 5x∧2 + 6x + 15 = x∧2 (2x+5) + 3 (2x+5) = (2x+5) 0 (x∧2+3)

God with you!!!

6 0

3 years ago

Read 2 more answers

In the triangle pictured, let A, B, C be the angles at the three vertices, and let a,b,c be the sides opposite those angles. Acc

Troyanec [42]

Answer:

Step-by-step explanation:

(a)

Consider the following:

$A=\frac{\pi}{4}=45°\\\\B=\frac{\pi}{3}=60°$

Use sine rule,

$\frac{b}{a}=\frac{\sinB}{\sin A}
\\\\=\frac{\sin{\frac{\pi}{3}}
}{\sin{\frac{\pi}{4}}}\\\\=\frac{[\frac{\sqrt{3}}{2}]}{\frac{1}{\sqrt{2}}}\\\\=\frac{\sqrt{2}}{2}\times \frac{\sqrt{2}}{1}=\sqrt{\frac{3}{2}}$

Again consider,

$\frac{b}{a}=\frac{\sin{B}}{\sin{A}}
\\\\\sin{B}=\frac{b}{a}\times \sin{A}\\\\\sin{B}=\sqrt{\frac{3}{2}}\sin {A}\\\\B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

Thus, the angle B is function of A is, $B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

Now find $\frac{dB}{dA}$

Differentiate implicitly the function $\sin{B}=\sqrt{\frac{3}{2}}\sin{A}$ with respect to A to get,

$\cos {B}.\frac{dB}{dA}=\sqrt{\frac{3}{2}}\cos A\\\\\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos A}{\cos B}$

b)

When $A=\frac{\pi}{4},B=\frac{\pi}{3}$ , the value of $\frac{dB}{dA}$ is,

$\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos {\frac{\pi}{4}}}{\cos {\frac{\pi}{3}}}\\\\=\sqrt{\frac{3}{2}}.\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}\\\\=\sqrt{3}$

c)

In general, the linear approximation at x= a is,

$f(x)=f'(x).(x-a)+f(a)$

Here the function $f(A)=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

At $A=\frac{\pi}{4}$

$f(\frac{\pi}{4})=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{\frac{\pi}{4}}]\\\\=\sin^{-1}[\sqrt{\frac{3}{2}}.\frac{1}{\sqrt{2}}]\\\\\=\sin^{-1}(\frac{\sqrt{2}}{2})\\\\=\frac{\pi}{3}$

And,

$f'(A)=\frac{dB}{dA}=\sqrt{3}$ from part b

Therefore, the linear approximation at $A=\frac{\pi}{4}$ is,

$f(x)=f'(A).(x-A)+f(A)\\\\=f'(\frac{\pi}{4}).(x-\frac{\pi}{4})+f(\frac{\pi}{4})\\\\=\sqrt{3}.[x-\frac{\pi}{4}]+\frac{\pi}{3}$

d)

Use part (c), when $A=46°$ , B is approximately,

$B=f(46°)=\sqrt{3}[46°-\frac{\pi}{4}]+\frac{\pi}{3}\\\\=\sqrt{3}(1°)+\frac{\pi}{3}\\\\=61.732°$

8 0

3 years ago