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3 years ago

Can some one help me on this the last answer since you can’t see it is CB=AC

Mathematics

1 answer:

Verdich [7]3 years ago

4 0

Answer:

C. AB/EF = BC/EF

Step-by-step explanation:

If the triangles are similar then the ratio of these side lengths must be equal.

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(2y^3+5y^2-y+15) ÷ (y-4)

Sveta_85 [38]

ANSWER: 2y^2 + 13y +51 r(219/y-4)

picture explanation too:

8 0

3 years ago

Tan x is ........ Cos x is ........ Sin x is ..........

fomenos

Answer:

Tan x= 3/4 Cos x= 4/5 Sin x=3/5

Step-by-step explanation:

"soh-cah-toa"

Just an easy way to remember how Sine, Cosine and Tangent work:

Soh...

Sine = Opposite / Hypotenuse

...cah...

Cosine = Adjacent / Hypotenuse

...toa

Tangent = Opposite / Adjacent

3 0

3 years ago

Read 2 more answers

PLEASEEE I NEED HELPPP!!!!

taurus [48]

Answer:

#carry on learning

mark me as brainliest

Find a point-slope form for the line that satisfies the stated conditions. Slope , passing through (-5,4)

I really need this question answered

By:

I don't see a value for the slope. We need that to set the equation, otherwise I can write an unlimited number of equations that pass through (-5,4).

I'll assume a slope so that you can see how the procedure would work. I like 6, so we'll assume a slope of 6.

The equation for a straight line has the form y = mx + b, where m is the slope and y is the y-intercept, the value of y when x = 0. We want a line that has slope 6, so:

y = 6x + b

We need to find b, so substitute the point (-5,4) that we know is on the line:

4 = 6*(-5) + b and solve for b

4 = -30 + b

b = 34

The line is y = 6x + 34

4 0

2 years ago

In a triangle ABC, measure of angle B is 90 degrees. AB is 3x-2 units and BC is x+3. If the area of the triangle is 17 sq cm, fo

Scorpion4ik [409]

Answer:

$x=\frac{8}{3}\ cm$

Step-by-step explanation:

we know that

The area of the right triangle ABC is equal to

$A=\frac{1}{2}(AB)(BC)$

we have

$A=17\ cm^2$

$AB=(3x-2)\ cm$

$BC=(x+3)\ cm$

substitute the values

$17=\frac{1}{2}(3x-2)(x+3)$

$34=(3x-2)(x+3)$

$34=3x^2+9x-2x-6$

$3x^2+7x-6-34=0$

$3x^2+7x-40=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$3x^2+7x-40=0$

$a=3\\b=7\\c=-40$

substitute in the formula

$x=\frac{-7(+/-)\sqrt{7^{2}-4(3)(-40)}} {2(3)}$

$x=\frac{-7(+/-)\sqrt{529}} {6}$

$x=\frac{-7(+/-)23} {6}$

$x=\frac{-7(+)23} {6}=\frac{16}{6}=\frac{8}{3}$

$x=\frac{-7(-)23} {6}=-5$

therefore

The solution is

$x=\frac{8}{3}\ cm$

6 0

3 years ago

What is 6,039 divided by 9 and please tell me if there a remainder

Alisiya [41]

The answer is 671
3,039/9
671
671*9
3,039

5 0

3 years ago

Read 2 more answers