All categories

3 years ago

What was the original price if: after increasing by 60% it became $8a?

Mathematics

2 answers:

mr Goodwill [35]3 years ago

7 0

Answer:

Step-by-step explanation:

We can make a mathematical equation with the information obtained:

<em>final value = initial value * (1 + percentage)</em>

So, if we want to know the <em>inicial value, </em>what we have to do is clear it.

So:

$initial value = \frac{final value}{1+percentage}$

So, in this case:

$initial value=\frac{8a}{1+0.60} = \frac{8a}{1.6}=5a$

Morgarella [4.7K]3 years ago

3 0

I think it’s 5......I think

You might be interested in

17. The perimeter of rectangle RPQT is 134dm.

Rina8888 [55]

Answer:

$\sf 1092 \: dm^2$

Refer to the attachment for explication of steps. :)

6 0

2 years ago

Write the number eighty seven in standard form

Alja [10]

Eighty seven written in standard form would be...
80 + 7 = 87

Good Luck! :)

4 0

3 years ago

Read 2 more answers

What is the Expanded Form for 27.39 using Powers Of 10's? Please help ASAP!!!

Licemer1 [7]

I think it's like 2.739 × 10

8 0

3 years ago

The diagram shows a logo made from 3 circles.

Cloud [144]

Answer:

% area of the shaded region = 33%

Step-by-step explanation:

Radius of the middle circle = 7 cm

Radius of the inner circle = 4 cm

Area of the middle circle = πr²

= π(7)²

= 49π

Area of the inner circle = π(4)²

= 16π

Area of the shaded region= 49π - 16π

= 33π

Area of the outermost circle = π(10)²

= 100π

% area of the shaded region = $\frac{\text{Area of the shaded region}}{\text{Area of the inner circle}}\times 100$

= $\frac{33\pi }{100\pi }\times 100$

= 33%

Therefore, area of the shaded region is 33% of the complete logo.

3 0

3 years ago

PLS ANSWER ASAP 30 POINTS!!! CHECK PHOTO! WILL MARK BRAINLIEST TO WHO ANSWERS

Sveta_85 [38]

I'll do Problem 8 to get you started

a = 4 and c = 7 are the two given sides

Use these values in the pythagorean theorem to find side b

$a^2 + b^2 = c^2\\\\4^2 + b^2 = 7^2\\\\16 + b^2 = 49\\\\b^2 = 49 - 16\\\\b^2 = 33\\\\b = \sqrt{33}\\\\$

With respect to reference angle A, we have:

opposite side = a = 4
adjacent side = b = $\sqrt{33}$
hypotenuse = c = 7

Now let's compute the 6 trig ratios for the angle A.

We'll start with the sine ratio which is opposite over hypotenuse.

$\sin(\text{angle}) = \frac{\text{opposite}}{\text{hypotenuse}}\\\\\sin(A) = \frac{a}{c}\\\\\sin(A) = \frac{4}{7}\\\\$

Then cosine which is adjacent over hypotenuse

$\cos(\text{angle}) = \frac{\text{adjacent}}{\text{hypotenuse}}\\\\\cos(A) = \frac{b}{c}\\\\\cos(A) = \frac{\sqrt{33}}{7}\\\\$

Tangent is the ratio of opposite over adjacent

$\tan(\text{angle}) = \frac{\text{opposite}}{\text{adjacent}}\\\\\tan(A) = \frac{a}{b}\\\\\tan(A) = \frac{4}{\sqrt{33}}\\\\\tan(A) = \frac{4\sqrt{33}}{\sqrt{33}*\sqrt{33}}\\\\\tan(A) = \frac{4\sqrt{33}}{(\sqrt{33})^2}\\\\\tan(A) = \frac{4\sqrt{33}}{33}\\\\$

Rationalizing the denominator may be optional, so I would ask your teacher for clarification.

So far we've taken care of 3 trig functions. The remaining 3 are reciprocals of the ones mentioned so far.

cosecant, abbreviated as csc, is the reciprocal of sine
secant, abbreviated as sec, is the reciprocal of cosine
cotangent, abbreviated as cot, is the reciprocal of tangent

So we'll flip the fraction of each like so:

$\csc(\text{angle}) = \frac{\text{hypotenuse}}{\text{opposite}} \ \text{ ... reciprocal of sine}\\\\\csc(A) = \frac{c}{a}\\\\\csc(A) = \frac{7}{4}\\\\\sec(\text{angle}) = \frac{\text{hypotenuse}}{\text{adjacent}} \ \text{ ... reciprocal of cosine}\\\\\sec(A) = \frac{c}{b}\\\\\sec(A) = \frac{7}{\sqrt{33}} = \frac{7\sqrt{33}}{33}\\\\\cot(\text{angle}) = \frac{\text{adjacent}}{\text{opposite}} \ \text{ ... reciprocal of tangent}\\\\\cot(A) = \frac{b}{a}\\\\\cot(A) = \frac{\sqrt{33}}{4}\\\\$

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

Summary:

The missing side is $b = \sqrt{33}$

The 6 trig functions have these results

$\sin(A) = \frac{4}{7}\\\\\cos(A) = \frac{\sqrt{33}}{7}\\\\\tan(A) = \frac{4}{\sqrt{33}} = \frac{4\sqrt{33}}{33}\\\\\csc(A) = \frac{7}{4}\\\\\sec(A) = \frac{7}{\sqrt{33}} = \frac{7\sqrt{33}}{33}\\\\\cot(A) = \frac{\sqrt{33}}{4}\\\\$

Rationalizing the denominator may be optional, but I would ask your teacher to be sure.

7 0

1 year ago