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2 years ago

Which equation describes the relationships

Mathematics

2 answers:

rosijanka [135]2 years ago

4 0

Answer:

Option A

Step-by-step explanation:

Sin(x) is the same value as cos(y) in the graph.

Phantasy [73]2 years ago

4 0

The answer is A just trust

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Find the surface area of the cylinder. Type an exact answer in terms of pi.

Elodia [21]

Answer:

=150.79644...cm²

~150cm²

Step-by-step explanation:

cylinder surface area = 2 pi r h + 2 pi r²

5 0

1 year ago

Read 2 more answers

Solve, please so it fast

jek_recluse [69]

I Think the answer is 17

6 0

2 years ago

The perimeter of a rectangle is 42 inches, and its area is 110 square inches. find the length and width of the rectangle.

lorasvet [3.4K]

$p = 2w + 2l$ $p = 42$
$a = l * w$ $a = 108$

You want to make either the length or width variable (I chose the width variable to be by itself. It doesn't matter) by itself by:

Divide both sides by 2.
(p) ÷ 2 = (2w + 2l) ÷ 2
$2's$ $cancel$ $out$
$\frac{p}{2} = w + l$

Subtract both sides by either length or width depending on what variable you chose to make by itself (in this case I subtracted the length variable because I wanted the width variable by itself).
$\frac{p}{2} - l = w + l - l$
$l's$ $cancel$ $out$
p/2 - l = w

$a = l * w$

You replace w with the equation above p/2 - l = w
$a = l( \frac{p}{2} - l)$

Multiply out the L.
$a = \frac{p}{2}l - l^{2}$

Plug in all your numbers a = 110 and p = 42
$110 = \frac{42}{2}l - l^{2}$
$110 = 21l - l^{2$

Move everything to the left side so $l^{2}$ would be positive (makes the equation easier when $l^{2}$ is positive).
$l^{2} - 21l + 110 = 0$

Factor.
$(l -10)(l-11) = 0$

Make each parenthesis set equal to 0.
$l-11=0$ $l-10=0$

Add.
$l = 11$ $l = 10$

By doing this you solve for both the width and length so the answer is:
w = 11 L = 10

8 0

3 years ago

A girl ran from her home to a store at the rate of 6 mph, and she walked back home at the rate of 4 mph. if it took 15 minutes f

viva [34]

Home is 0.6 miles away from the store

How to solve such time related questions?

The key concept for solving such questions is the relationship between speed, distance and time.

The relationship between them is Distance = Speed x Time

Given:-

Total time taken by girl to complete the trip = 15 minutes = $\frac{1}{4}$ hours

Speed while riding going = 6 miles per hour

Speed while coming back = 4 miles per hour

Total time taken by girl =

Time taken while going + Time taken while coming back - (1)

Let x be distance from home to store

$\frac{x}{6}$ = Time taken while going

$\frac{x}{4}$ = Time taken while coming back

Substituting in (1) we get

$\frac{1}{4} = \frac{x}{6} + \frac{x}{4}$

$\frac{1}{4} = \frac{2x + 3x}{12}$

$\frac{1}{4} = \frac{5x}{12}$

$x= \frac{12}{20}$

x = 0.6 miles

Thus home is 0.6 miles away from the store

Learn more about time related here :

brainly.com/question/12759408

#SPJ4

6 0

1 year ago

[30 POINTS] Please help!!!

Len [333]

Answer:

Part 1) $y=1.5x+5$

Part 2) $y=-(2/3)x-(11/3)$

Part 3) $y=0.25x+2.75$

Part 4) $y=-2x+5$

Part 5) $y=0.5x-1$

Part 6) The graph in the attached figure

Step-by-step explanation:

Part 1) we have

$m=3/2=1.5$

$point(-2,2)$

The equation of the line into point slope form is equal to

$y-y1=m(x-x1)$

substitute

$y-2=1.5(x+2)$

$y=1.5x+3+2$

$y=1.5x+5$

Part 2) we know that

If two lines are perpendicular

then

the product of their slopes is equal to minus one

$m1*m2=-1$

the slope of the line 1 is equal to

$m1=1.5$

Find the slope m2

$1.5*m2=-1$

$m2=-2/3$

Find the equation of the line 2

we have

$m2=-2/3$

$point(-7,1)$

The equation of the line into point slope form is equal to

$y-y1=m(x-x1)$

substitute

$y-1=(-2/3)(x+7)$

$y=-(2/3)x-(14/3)+1$

$y=-(2/3)x-(11/3)$

Part 3) we have

$m=1/4=0.25$

$point(1,3)$

The equation of the line into point slope form is equal to

$y-y1=m(x-x1)$

substitute

$y-3=0.25(x-1)$

$y=0.25x-0.25+3$

$y=0.25x+2.75$

Part 4) we have

$m=-2$

$b=5$ -----> y-intercept

we know that

The equation of the line into slope intercept form is equal to

$y=mx+b$

substitute the values

$y=-2x+5$

Part 5) we have that

The slope of the line 4 is equal to $-2$

the slope of the line perpendicular to the line 4 is equal to

$-2*m=-1\\m=(1/2)=0.5$

therefore

in this problem we have

$m=0.5$

$point(-2,-2)$

The equation of the line into point slope form is equal to

$y-y1=m(x-x1)$

substitute

$y+2=0.5(x+2)$

$y=0.5x+1-2$

$y=0.5x-1$

Part 6)

using a graphing tool

see the attached figure

3 0

2 years ago