Helllpppp meee if you answer in 2 minutes you get brainliest !!!!!!!!!!!!!!!!

Please please help! Ill give brainliest <br> worded problem image attached

Ulleksa [173]

The equation is:

I=100+12J+5T+7S+5H-8M

The given information is that I=197, J=5, T=5, H=3, and M=3, and we’re trying to solve for the amount of shorts (S) he sold.

We can plug in the variables into the equation:

(197)=100+12(5)+5(5)+7S+5(3)-8(3)

After simplification:

197=100+60+25+7S+15-24

We can add/subtract the constants:

197=100+60+25+7S+15-24

We will add/subtract the constants:

197=176+7S

We will subtract 176 from both sides of the equation to cancel it out and isolate 7S:

197-176=176-176+7S

21=7S

We will divide 7 from both sides of the equation to cancel it out and isolate S:

21/7=7S/7

3=S or S=3 with the symmetric property of Equality applied.

Therefore, he sold 3 pairs of shorts

4 0

3 years ago

Find Mx, My, and (x, y) for the lamina of uniform density rho bounded by the graphs of the equations. y = x2/3, y = 0, x = 1

erik [133]

Answer:

$\mathbf{(\overline x , \overline y ) = (\dfrac{5}{8}, \dfrac{5}{14})}$

Step-by-step explanation:

Given that:

$y = x^{2/3}$ at y = 0 , x = 1

Then:

Area = $\int^{1}_{0} x^{2/3} \ dx$

Area = $\begin {bmatrix} \dfrac{3}{5}x^{5/3} \end {bmatrix} ^1_0$

Area = $\dfrac{3}{5}$

Then:

$\overline x = \dfrac{1}{A} \int^b_a x (f(x) -g(x) ) \ dx$

$\overline x = \dfrac{5}{3} \int^1_0 x (x^{2/3} -0 ) \ dx$

$\overline x = \dfrac{5}{3} \int^1_0 x^{5/3} \ dx$

$\overline x = \dfrac{5}{3} \ [\dfrac{3}{8}x^{8/3}]^1_0$

$\overline x = \dfrac{5}{3} \times \dfrac{3}{8}$

$\overline x = \dfrac{5}{8}$

Similarly;

$\overline y = \dfrac{1}{A} \int^b_a \dfrac{1}{2} \begin{bmatrix} (f(x)^)2 - (g(x))^2 \end {bmatrix} \ dx$

$\overline y = \dfrac{5}{3} \int^1_0 \dfrac{1}{2} \begin{bmatrix} (f(x^{2/3})^2 -0 \end {bmatrix} \ dx$

$\overline y = \dfrac{5}{3} \int^1_0 \dfrac{1}{2} \begin{bmatrix} (x^{4/3} ) \end {bmatrix} \ dx$

$\overline y = \dfrac{5}{3} \begin{bmatrix} \dfrac{1}{2} (x^{7/3} ) \times \dfrac{3}{7} \end {bmatrix} ^1_0$

$\overline y = \dfrac{5}{3} \begin{bmatrix} \dfrac{3}{14} (x^{7/3} ) \end {bmatrix} ^1_0$

$\overline y = (\dfrac{5}{3} \times \dfrac{3}{14} )$

$\overline y = \dfrac{5}{14}$

Thus; $\mathbf{(\overline x , \overline y ) = (\dfrac{5}{8}, \dfrac{5}{14})}$

4 0

3 years ago

The length of a rectangular mural is 2 ft more than three times the height. The area is 165 ft^2. Find the height of the mural.

NISA [10]

Answer:

x=-1±4√31/3 ft is the height

Step-by-step explanation:

let the height be x ft.

then the length would be 3x+2

area = L*H= x(3x+2)

165=3x²+2x

0= 3x²+2x -165

dividing the whole equation by 3

0=x²+2x/3-55

(x)²+2(x)(1/3) +(1/3)² -55 -(1/9)=0

(x+1/3)²- 495-1/9=0

(x+1/3)²= 496/9

taking square root on both sides

(x+1/3)= ±4√31/3

x= ±4√31/3-1/3

x=-1±4√31/3 ft is the height

5 0

4 years ago

Help me please I need it by tommrow u get brai nlist

lakkis [162]

Answer:

92905

Step-by-step explanation:

27,500=0.296e

29.6%=0.296

27,500/0.296 =e

e rounded to the nearest whole number is 92,905

4 0

3 years ago

the number of girls on the school's basketball team decreased by 20% this year. if there were 25 girls on the team last year. ho

Harman [31]

Answer:

Step-by-step explanation:

okay so to get 10% of 20 you just move the decimal to the right one so 25.0 would become 2.5 then to find 20% of that you multiply by 2 so 2.5 x 2 = 5 so that means the team lost 5 girls. 25-5=20 therefor there are 20 players on the team this year

4 0

3 years ago