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3 years ago

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4. Mass is measured using a

1 answer:

aniked [119]3 years ago

3 0

Correct Answer: Scale

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Please help will give brainliest<br> options are <br> w=v/lh<br> w=lh/v<br> w=vl/h<br> w=vm/l

RUDIKE [14]

Answer:

B

Step-by-step explanation:

7 0

3 years ago

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last week, Jim worked h hours and earned $9 per hour. if he earned a total of $405, which equation can be used to find out how m

ad-work [718]

I hope this helps you

if he can earn $9 per hour

$405 ? hours

?=405/9

?=45 hours he worked

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3 years ago

A persons resting heart rate is typically between 60 and 100 beats per minute. Noah looks at his watch, and counts 8 heartbeats

gulaghasi [49]

Its typical because 60 between 100. 8 times 10 seconds equals 80 which is in between so Noah's heart rate is typical.

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3 years ago

The box plots show the data distributions for the number of text messages Will and Jaime sent each day.What is the difference of

Softa [21]

Hey there!

You can find the medians by referring to the line in the middle of both of the boxes. These indicate the median of the sets of data you have. The first median is at 10 and the other is at 12.

All you need to do to find the difference is subtract the smaller number from the bigger number.

12 – 10 = 2

Your difference will be 2, making your second option your answer. Hope this helped you out! :-)

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3 years ago

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Cosθ=−2√3 , where π≤θ≤3π2 .

Alex787 [66]

Answer:

$sin(\theta + \beta) = -\frac{\sqrt{7}}{5}-4\frac{\sqrt{2}}{15}$

Step-by-step explanation:

step 1

Find the $sin(\theta)$

we know that

Applying the trigonometric identity

$sin^2(\theta)+ cos^2(\theta)=1$

we have

$cos(\theta)=-\frac{\sqrt{2}}{3}$

substitute

$sin^2(\theta)+ (-\frac{\sqrt{2}}{3})^2=1$

$sin^2(\theta)+ \frac{2}{9}=1$

$sin^2(\theta)=1- \frac{2}{9}$

$sin^2(\theta)= \frac{7}{9}$

$sin(\theta)=\pm\frac{\sqrt{7}}{3}$

Remember that

π≤θ≤3π/2

so

Angle θ belong to the III Quadrant

That means ----> The sin(θ) is negative

$sin(\theta)=-\frac{\sqrt{7}}{3}$

step 2

Find the sec(β)

Applying the trigonometric identity

$tan^2(\beta)+1= sec^2(\beta)$

we have

$tan(\beta)=\frac{4}{3}$

substitute

$(\frac{4}{3})^2+1= sec^2(\beta)$

$\frac{16}{9}+1= sec^2(\beta)$

$sec^2(\beta)=\frac{25}{9}$

$sec(\beta)=\pm\frac{5}{3}$

we know

0≤β≤π/2 ----> II Quadrant

so

sec(β), sin(β) and cos(β) are positive

$sec(\beta)=\frac{5}{3}$

Remember that

$sec(\beta)=\frac{1}{cos(\beta)}$

therefore

$cos(\beta)=\frac{3}{5}$

step 3

Find the sin(β)

we know that

$tan(\beta)=\frac{sin(\beta)}{cos(\beta)}$

we have

$tan(\beta)=\frac{4}{3}$

$cos(\beta)=\frac{3}{5}$

substitute

$(4/3)=\frac{sin(\beta)}{(3/5)}$

therefore

$sin(\beta)=\frac{4}{5}$

step 4

Find sin(θ+β)

we know that

$sin(A + B) = sin A cos B + cos A sin B$

so

In this problem

$sin(\theta + \beta) = sin(\theta)cos(\beta)+ cos(\theta)sin (\beta)$

we have

$sin(\theta)=-\frac{\sqrt{7}}{3}$

$cos(\theta)=-\frac{\sqrt{2}}{3}$

$sin(\beta)=\frac{4}{5}$

$cos(\beta)=\frac{3}{5}$

substitute the given values in the formula

$sin(\theta + \beta) = (-\frac{\sqrt{7}}{3})(\frac{3}{5})+ (-\frac{\sqrt{2}}{3})(\frac{4}{5})$

$sin(\theta + \beta) = (-3\frac{\sqrt{7}}{15})+ (-4\frac{\sqrt{2}}{15})$

$sin(\theta + \beta) = -\frac{\sqrt{7}}{5}-4\frac{\sqrt{2}}{15}$

8 0

3 years ago