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3 years ago

PLEASE, PLEASE, PLEASE I NEED HELP IT TEN POINT AND WILL BE MARK AS BRAINLIST! ttttttthhhhhhhaaaaaannnnnnkkkk

Mathematics

1 answer:

kow [346]3 years ago

6 0

Answer: B.1/6 quarts

Step-by-step explanation:

3/4 Minutes = 1/8 Juice

4/3 * 3/4 Minutes * 4/3 * 1/8 Juice

1 Minutes = 1/6 quarts

Making B your answer

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A prize of $100,$200,$500, or $1000 will be randomly awarded to a game show contestant, with the probabilities of winning each p

guajiro [1.7K]

Answer: 400

Step-by-step explanation:

A P E X

4 0

3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWERS ONLY PLEASE!!!

Kisachek [45]

Answer: (A) vertical asymptote: x = 2, horizontal asymptote: y = 1

Step-by-step explanation:

$f(x) = \dfrac{1}{x-2}+1$

Vertical Asymptote is the restriction on the x-value. The denominator cannot be zero, so x - 2 ≠ 0 ⇒ x ≠ 2

The restricted value on x is when x = 2 which is the vertical asymptote

Horizontal Asymptote (H.A.) is the restriction on the y-value. This is a comparison of the numerator (n) and denominator (m). There are 3 rules that will help you:

n > m No H.A. (use long division to find slant asymptote)
n = m H.A. is the coefficient of n divided by coefficient of m
n < m H.A. is 0

In the given problem, n < m so y = 0, however there is also a vertical shift of up 1 so the H.A. also shifts up. This results in H.A. of y = 1

5 0

3 years ago

Help pleasee its due at 8:00 pm

bogdanovich [222]

Answer:

10L

Step-by-step explanation:

4 0

3 years ago

For the figures below, assume they are made of semicircles, quarter circles and squares. For each shape, find the area and perim

Marina86 [1]

Area of the shaded region $$=36(\pi -2)$ square cm

Perimeter of the shaded region $=6 (\pi + 2\sqrt 2)$ cm

Solution:

Radius of the quarter of circle = 12 cm

Area of the shaded region = Area of quarter of circle – Area of the triangle

$$=\frac{1}{4} \pi r^2 - \frac{1}{2} bh$

$$=\frac{1}{4} \pi \times 12^2 - \frac{1}{2} \times 12 \times 12$

$$=36\pi -72$

$$=36(\pi -2)$ square cm.

Area of the shaded region $$=36(\pi -2)$ square cm

Using Pythagoras theorem,

$AC^2=AB^2+BC^2$

$AC^2=12^2+12^2$

$AC^2=288$

Taking square root on both sides of the equation, we get

$AC= 12\sqrt 2$ cm

Perimeter of the quadrant of a circle = $\frac{1}{4} \times 2\pi r$

$$=\frac{1}{4} \times 2 \times \pi \times 12$

$$=6 \pi$ cm

Perimeter of the shaded region = $6 \pi + 12\sqrt 2$ cm

$=6 (\pi + 2\sqrt 2)$ cm

Hence area of the shaded region $$=36(\pi -2)$ square cm

Perimeter of the shaded region $=6 (\pi + 2\sqrt 2)$ cm

6 0

2 years ago

19x+15y=-6 changed into slope intercept form.

attashe74 [19]

Answer:

y = - $\frac{19}{15}$ x - $\frac{2}{5}$

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Given

19x + 15y = - 6 ( subtract 19x from both sides )

15y = - 19x - 6 ( divide all terms by 15 )

y = - $\frac{19}{15}$ x - $\frac{6}{15}$ , that is

y = - $\frac{19}{15}$ x - $\frac{2}{5}$ ← in slope- intercept form

7 0

3 years ago

Read 2 more answers