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3 years ago

Helppppp mathhhhhhhhhhhh

Mathematics

2 answers:

Karo-lina-s [1.5K]3 years ago

8 0

Y - 3 = 2x -1 ...y = 2x + 2

x +3 = y - 1

replace y = 2x + 2 into x +3 = y - 1

x +3 = 2x + 2 - 1
2x -x = 3 -1
x = 2

y = 2x + 2
y = 2(2) + 2 = 6

answer x = 2 , y = 6 first choice

Basile [38]3 years ago

5 0

Try to substitute the values of X and Y inside, and use Pythagorean theorem to see if the triangles are congruent.

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Need help on polynomials

kkurt [141]

Follow these steps and you should get your answer its not the same questions but it should help
The answer to the problem is as follows:

These two polynomials are 5x^2 + 3x + 5 and 3x^2 +2x + 2:

Subtracting the second expression from the first:

<span>5x^2 + 3x + 5
- (3x^2 +2x + 2)
</span>____________
<span>2x2 + x + 3 <------------- The difference
</span>

7 0

3 years ago

What is the product of 1 x 1 and 5x2 2x 6 ? Write your answer in standard form. 24

Svetradugi [14.3K]

Answer:

1. 1

2. 10

3. 12

Step-by-step explanation:

8 0

3 years ago

If one face of the cube has an area of 25cm calculatethe total surface area

DIA [1.3K]

there are 6 faces multiply 25 by 6

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3 0

3 years ago

If A and B are mutually exclusive events with P(A)= 0.3 and P(B)= 0.5, then P(A and B)=

arsen [322]

Answer:

$P (A\ and\ B) = 0$

Step-by-step explanation:

By definition if two events are mutually exclusive then they are disjoint events, that is, they can not occur at the same time.

In general terms we say that events are mutually exclusive, then:

$P (A\ and\ B) = 0$

In this case we have that

$P(A)= 0.3\\\\P(B)= 0.5$

But A and B are mutually exclusive events, then $P (A\ and\ B) = 0$

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