All categories

3 years ago

Helpppp and explain this word problem!!!

Mathematics

1 answer:

Mashutka [201]3 years ago

6 0

Answer: 90 points (second option)

500+2x/8=85
500+2x=680
2x=180
x=90
90 on each test.

You might be interested in

What is the slope of the line that contains the points (-4, 2) and (6, -3)?

mote1985 [20]

Answer:

-1/2 or $-\frac{1}{2}$

Step-by-step explanation:

to find the slope, use the formula

(y₁ - y₂)/(x₁ - x₂)

(2 - -3)/(-4 - 6)

(2 + 3)/(- 4 - 6)

5/-10

1/-2

-1/2

8 0

3 years ago

Does anyone know this one I would really need help

Novay_Z [31]

Answer:

Step-by-step explanation:

ok so first you eat my poop then you eat your own poop and then you eat my dog poop

7 0

3 years ago

Juanita will work for 4 7/8 hours at her job on Saturday. if she also volunteers for 2 1/3 hours at the hospital, how many hours

zheka24 [161]

Answer:

I think it's D...maybe

Step-by-step explanation:

8 0

3 years ago

PLS HELP ME I CANT DO IT AND JMBEING TIMED

tamaranim1 [39]

Answer:

For A+B

a. 0.3+0.8 = 1.1

b. 1.2+1.4 = 2.6

c. 3.4 + 4.2 = 7.6

d. 2.25 + 2.95 = 5.20

For B-A

a. 0.8-0.3 = 0.5

b. 1.4-1.2 = 0.2

c. 4.2-3.4 = 1.2

d. 2.95-2.25 = .070

8 0

3 years ago

find the area of the trapezium whose parallel sides are 25 cm and 13 cm The Other sides of a Trapezium are 15 cm and 15 CM

Snezhnost [94]

$\huge\underline{\red{A}\blue{n}\pink{s}\purple{w}\orange{e}\green{r} -}$

Given - A trapezium ABCD with non parallel sides of measure 15 cm each ! along , the parallel sides are of measure 13 cm and 25 cm

To find - Area of trapezium

Refer the figure attached ~

In the given figure ,

AB = 25 cm

BC = AD = 15 cm

CD = 13 cm

Construction -

$draw \: CE \: \parallel \: AD \: \\ and \: CD \: \perp \: AE$

Now , we can clearly see that AECD is a parallelogram !

$\therefore$ AE = CD = 13 cm

Now ,

$AB = AE + BE \\\implies \: BE =AB - AE \\ \implies \: BE = 25 - 13 \\ \implies \: BE = 12 \: cm$

Now , In ∆ BCE ,

$semi \: perimeter \: (s) = \frac{15 + 15 + 12}{2} \\ \\ \implies \: s = \frac{42}{2} = 21 \: cm$

Now , by Heron's formula

$area \: of \: \triangle \: BCE = \sqrt{s(s - a)(s - b)(s - c)} \\ \implies \sqrt{21(21 - 15)(21 - 15)(21 - 12)} \\ \implies \: 21 \times 6 \times 6 \times 9 \\ \implies \: 12 \sqrt{21} \: cm {}^{2}$

Also ,

$area \: of \: \triangle \: = \frac{1}{2} \times base \times height \\ \\\implies 18 \sqrt{21} = \: \frac{1}{\cancel2} \times \cancel12 \times height \\ \\ \implies \: 18 \sqrt{21} = 6 \times height \\ \\ \implies \: height = \frac{\cancel{18} \sqrt{21} }{ \cancel 6} \\ \\ \implies \: height = 3 \sqrt{21} \: cm {}^{2}$

Since we've obtained the height now , we can easily find out the area of trapezium !

$Area \: of \: trapezium = \frac{1}{2} \times(sum \: of \:parallel \: sides) \times height \\ \\ \implies \: \frac{1}{2} \times (25 + 13) \times 3 \sqrt{21} \\ \\ \implies \: \frac{1}{\cancel2} \times \cancel{38 }\times 3 \sqrt{21} \\ \\ \implies \: 19 \times 3 \sqrt{21} \: cm {}^{2} \\ \\ \implies \: 57 \sqrt{21} \: cm {}^{2}$

hope helpful :D

6 0

2 years ago