All categories

3 years ago

Increase 53.20 by 7%

Mathematics

2 answers:

____ [38]3 years ago

7 0

Answer:

56.924

Step-by-step explanation:

Lapatulllka [165]3 years ago

3 0

Step-by-step explanation:

Change the problem into something you recognize

7% = 7 x 10^-2 = 0.07

Multiply by this decimal

53.20 x 0.07 = a

Then add this to the original

53.20 + a

You might be interested in

Express 0.002304 in standard form , correct to one decimal place

PSYCHO15rus [73]

Answer:

$= 2.3 \times {10}^{ - 3}$

6 0

3 years ago

Read 2 more answers

I need help with this please

enyata [817]

Answer:¿Qué es lo contrario de estos estadistas?

Step-by-step explanation:

What is the opposite of these statesment

5 0

3 years ago

I need help with finding the answer to a) and b). Thank you!

shtirl [24]

Answer:

$\displaystyle \sin\Big(\frac{x}{2}\Big) = \frac{7\sqrt{58} }{ 58 }$

$\displaystyle \cos\Big(\frac{x}{2}\Big)=-\frac{3 \sqrt{58}}{58}$

$\displaystyle \tan\Big(\frac{x}{2}\Big)=-\frac{7}{3}$

Step-by-step explanation:

We are given that:

$\displaystyle \sin(x)=-\frac{21}{29}$

Where x is in QIII.

First, recall that sine is the ratio of the opposite side to the hypotenuse. Therefore, the adjacent side is:

$a=\sqrt{29^2-21^2}=20$

So, with respect to x, the opposite side is 21, the adjacent side is 20, and the hypotenuse is 29.

Since x is in QIII, sine is negative, cosine is negative, and tangent is also negative.

And if x is in QIII, this means that:

$180$

So:

$\displaystyle 90 < \frac{x}{2} < 135$

Thus, x/2 will be in QII, where sine is positive, cosine is negative, and tangent is negative.

Recall that:

$\displaystyle \sin\Big(\frac{x}{2}\Big)=\pm\sqrt{\frac{1 - \cos(x)}{2}}$

Since x/2 is in QII, this will be positive.

Using the above information, cos(x) is -20/29. Therefore:

$\displaystyle \sin\Big(\frac{x}{2}\Big)=\sqrt{\frac{1 + 20/29}{2}$

Simplify:

$\displaystyle \sin\Big(\frac{x}{2}\Big)=\sqrt{\frac{49/29}{2}}=\sqrt{\frac{49}{58}}=\frac{7}{\sqrt{58}}=\frac{7\sqrt{58}}{58}$

Likewise:

$\displaystyle \cos \Big( \frac{x}{2} \Big) =\pm \sqrt{ \frac{1+\cos(x)}{2} }$

Since x/2 is in QII, this will be negative.

Using the above information, cos(x) is -20/29. Therefore:

$\displaystyle \cos \Big( \frac{x}{2} \Big) =-\sqrt{ \frac{1- 20/29}{2} }$

Simplify:

$\displaystyle \cos\Big(\frac{x}{2}\Big)=-\sqrt{\frac{9/29}{2}}=-\sqrt{\frac{9}{58}}=-\frac{3}{\sqrt{58}}=-\frac{3\sqrt{58}}{58}$

Finally:

$\displaystyle \tan\Big(\frac{x}{2}\Big) = \frac{\sin(x/2)}{\cos(x/2)}$

Therefore:

$\displaystyle \tan\Big(\frac{x}{2}\Big)=\frac{7\sqrt{58}/58}{-3\sqrt{58}/58}=-\frac{7}{3}$

5 0

3 years ago

What is the final answer

Anna35 [415]

Answer:

y=3x-2 (d)

Step-by-step explanation:

First, we need to look at the y-intercept, we can see that when x = 0, y= -2. So then our equation has to be y=mx+b, and b is -2. next we find our slope, we can see that we go up by 3, then over 1, so our slope is 3/1 or 3. so the final equation is y=3x-2.

Hope this helps!

8 0

3 years ago

Read 2 more answers

Pleaseeee 10 points for this answer

Sever21 [200]

Answer:

Leave the answer you’re looking for is C hope this helps

Step-by-step explanation:

6 0

4 years ago

Read 2 more answers