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3 years ago

What is the solution for x in the equation. 4x - 3+5 = 2x + 7 - 8x

Mathematics

1 answer:

FrozenT [24]3 years ago

5 0

Answer:

hence solution for x in the equation is x= 1/2

Step-by-step explanation:

4x - 3+5 = 2x + 7 - 8x

4x +2 = -6x +7

taking x to the other side then we get

4x+6x = -2+7

10x = 5

divide both side by 10

10x/10 =5/10

x= 1/2

hence solution for x in the equation is x= 1/2

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Determine the greatest common factor (GCF): 10, 20, 40 Cual es la respuesta correcta

aniked [119]

Answer:

GCF would be 10

Step-by-step explanation:

GCF, aka the greatest common factor is the biggest number that can be multiplied to get that number. for example, in this situation it would be 1- because 10 can be multiplied to get any of those numbers. 1 x 10 = 10 ; 2 x 10 =20 ; 4 x 10 = 40. hope this helped!!

4 0

3 years ago

Amy buys 8 sandwiches and 12 waters.her total cost is 8s+12 how much did she pay. S=sandwiches

Mrac [35]

Answer:

8s + 12

s=12-8

s=4

8s +12

s=12+8

s=20

Step-by-step explanation:

im not sure kung minus o plus kasi kung kinus mababa lang binayad nya

4 0

2 years ago

Read 2 more answers

What is the length of the hypotenuse? If necessary, round to the nearest tenth.

exis [7]

Answer:

Step-by-step explanation:

the length of the hypotenuse is 4 Km, it is probable that you are looking for the value of the cathetus a.

it is a right triangle and we use Pythagoras

a² = 4² - 2²

a² = 16 - 4

a² = 12

a = √12

a = 3.46 (round 3.5)

4 0

1 year ago

A) SSS B) SAS C) ASA D) AAS

amm1812

This is B.SAS (side angle side)

4 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