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2 years ago

Help me i need it to pass

Mathematics

1 answer:

Harrizon [31]2 years ago

5 0

The answer is 8.

8x4=32

(8x7)-(8x3)=32

Hope this helps!

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There are two non congruent triangles where B=55 degrees a=15 and b=13. Find the measures of the angles of the triangle with the

liq [111]

Answer:

The side c needs to be 12.85 for the triangle to have greatest perimeter

Step-by-step explanation:

We are given;

b = 13

a = 15

Angle at b: B = 55°

Let's find side c.

Using the law of cosines,we have;

b² = a² + c² - 2ac•cos(B)

13² = 15² + c² - 2•15•c•cos(55)

169 = 225 + c² - 30c•cos(55)

c² - 30c•cos(55) + 225 - 169 = 0

c² - 30c•cos(55) + 56 = 0

c² - 30c•(0.5736) + 56 = 0

c² - 17.208c + 56 = 0

Using quadratic formula;

c = [-(-17.208) ± √((-17.208)² - (4•1•56)]/2(1)

c = [17.208 ± √(296.115 - 224)]/2

c = 8.604 ± 4.246

To have the greater perimeter, we need the larger value of c, thus we will use the positive sign and ignore the negative one ;

Thus,

c = 8.604 + 4.246 = 12.85

5 0

2 years ago

What is 46785392+78 ?

worty [1.4K]

The answer is 46,785,470

7 0

2 years ago

Read 2 more answers

Wich is the first step in simplifying the expression 2(x+3)+5

34kurt

Mult 2 by x you'll get 2x then 2 by 3 you'll get 6 and add 6 to 5, after that you'll get 11 it should look like this 2x+11.

7 0

3 years ago

Read 2 more answers

Help me please please please PLEASE! NO LINKS

tresset_1 [31]

Answer:

for the first one its right one down 4

for the second one its right 4 down 1

for the last one its left 1 up 4

im pretty sure i helped u with your other one too

hope this helps

7 0

2 years ago

Read 2 more answers

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

2 years ago