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3 years ago

Help me please I’ll mark brainliest thank you

Mathematics

1 answer:

jenyasd209 [6]3 years ago

4 0

Answer:

11/15=x/240

240/15=16

11 x 16=176

11/15=176/240 or rather he completes 176 passes

Step-by-step explanation:

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Please help!! I can’t figure this out

Rudiy27

Each room has 2 walls that are 4x10 = 40 square feet and 2 walls that are 8x10 = 80 square feet. So 80+80+40+40 = 240 square feet of wall will be painted per room.

We use 2 coats of paint so in effect, we double the area. Imagine laying one thin wall on top of the other. The amount of paint area used per room doubles from 240 to 480

There are 4 of these rooms so 4*480 = 1920

The total area we need to paint is 1920 square feet

Divide this over 32 to get 1920/32 = 60

Final Answer: 60

5 0

3 years ago

What are the ratios of sine, cosine, and tangent for angle Y? sin(Y) = StartFraction X Z Over X Y EndFraction; cos(Y) = StartFra

Sholpan [36]

Answer:

$\sin Y= \frac{XZ}{XY}$

$\cos Y= \frac{YZ}{XY}$

$\tan Y= \frac{XZ}{YZ}$

Step-by-step explanation:

Given

See attachment for triangle

Required

Find $\sin, \cos$ and $\tan$ of angle Y

For angle Y:

$Opposite = XZ$

$Adjacent = YZ$

$Hypotenuse = XY$

The $\sin$ of an angle is calculated as:

$\sin\theta = \frac{Opposite}{Hypotenuse}$

So:

$\sin Y= \frac{XZ}{XY}$

The $\cos$ of an angle is calculated as:

$\cos\theta = \frac{Adjacent}{Hypotenuse}$

So:

$\cos Y= \frac{YZ}{XY}$

The $\tan$ of an angle is calculated as:

$\tan\theta = \frac{Opposite}{Adjacent}$

So:

$\tan Y= \frac{XZ}{YZ}$

7 0

3 years ago

Read 2 more answers

Explain how a bar model can be used to show a subtraction problem

snow_lady [41]

The amount of money you have each month will be subtracted by the amount that you spend then you would put amount you spent on a bar graph alond with the amount you had

6 0

3 years ago

Which point is in the solution set of the given system of inequalities 3x+y>-3,x+2y<4

Dima020 [189]

Step 1. Solve both inequalities for $y$ :
$3x+y\ \textgreater \ -3$
$y\ \textgreater \ -3x-3$

$x+2y\ \textless \ 4$
$2y\ \textless \ -x+4$
$y\ \textless \ - \frac{1}{2} x+2$

Step 2. To check a point in the solution of the given system of inequalities, look for the intercepts of the lines $-3x-3$ and $- \frac{1}{2} x+2$ :

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

Replace (1) in (2):
$-3x-3=- \frac{1}{2} x+2$
Solve for $x$ :
$\frac{5}{2} x=-5$
$x=-2$ (3)

Replace (3) in (1):
$y=-3x-3$
$y=-3(-2)-3$
$y=6-3$
$y=3$

We can conclude that the point (-2,3) is in the solution of the system if <span>inequalities</span>; also any point inside the dark shaded area of the graph of the system of inequalities is also a solution of the system.