Sheri's freezer is 2 feet wide 6feet long and 2 feet deep what's the volume of her freeser

A recipe for sparkling grape juice calls for 1/2 quarts of sparkling water and 3/4 quarts of grape juice how much sparkling wate

dimaraw [331]

Answer: 6

Step-by-step explanation:

3/4 x 12 =36/4

12 x 1/2=6

6 0

3 years ago

What is the value of x? 40,000 mg = 10x g. A.400 B.40 C.4 or D 0.4

MissTica

I hope this helps you

3 0

3 years ago

Read 2 more answers

When added, what is the simplified form of (7 + 2i) + (3 + 6i)? Write the answer in standard form, a + bi, where a and b are rea

Liula [17]

9514 1404 393

Answer:

10 +8i

Step-by-step explanation:

You add these the same way you would if they were considered to be polynomials.

(7 +2i) +(3 +6i) = (7 +3) +(2 +6)i = 10 +8i

5 0

3 years ago

Mrs. Mack teaches three art classes, and there are 30 students in each dass. She has a supply

laila [671]

Answer: 3/5

Step-by-step explanation:

1. Multiply 30 by 3 (90)

2. 90 becomes the numerator and 150 becomes the denominator

3. Divide the numerator (90) and the denominator (150) by the greatest common factor (30), to simplify the fraction

4. You should come to the answer 3/5

Hope this helps.

6 0

3 years ago

An equation of a hyperbola is given.

siniylev [52]

Answer:

a)

The vertices are $\left(3,\:0\right),\:\left(-3,\:0\right)$ .

The foci are $\left(3\sqrt{5},\:0\right),\:\left(-3\sqrt{5},\:0\right)$ .

The asymptotes are $y=2x,\:y=-2x$ .

b) The length of the transverse axis is 6.

c) See below.

Step-by-step explanation:

$\frac{\left(x-h\right)^2}{a^2}-\frac{\left(y-k\right)^2}{b^2}=1$ is the standard equation for a right-left facing hyperbola with center $\left(h,\:k\right)$ .

a)

The vertices $\:\left(h+a,\:k\right),\:\left(h-a,\:k\right)$ are the two bending points of the hyperbola with center $\:\left(h,\:k\right)$ and semi-axis a, b.

Therefore,

$\frac{x^2}{9}-\frac{y^2}{36}=1$ , is a right-left Hyperbola with $\:\left(h,\:k\right)=\left(0,\:0\right),\:a=3,\:b=6$ and vertices $\left(3,\:0\right),\:\left(-3,\:0\right)$ .

For a right-left facing hyperbola, the Foci (focus points) are defined as $\left(h+c,\:k\right),\:\left(h-c,\:k\right)$ where $c=\sqrt{a^2+b^2}$ is the distance from the center $\left(h,\:k\right)$ to a focus.

Therefore,

$\frac{x^2}{9}-\frac{y^2}{36}=1$ , is a right-left Hyperbola with $\:\left(h,\:k\right)=\left(0,\:0\right),\:a=3,\:b=6$ $c=\sqrt{3^2+6^2}= 3\sqrt{5}$ and foci $\left(3\sqrt{5},\:0\right),\:\left(-3\sqrt{5},\:0\right)$

The asymptotes are the lines the hyperbola tends to at $\pm \infty$ . For right-left hyperbola the asymptotes are: $y=\pm \frac{b}{a}\left(x-h\right)+k$

Therefore,

$\frac{x^2}{9}-\frac{y^2}{36}=1$ , is a right-left Hyperbola with $\:\left(h,\:k\right)=\left(0,\:0\right),\:a=3,\:b=6$ and asymptotes

$y=\frac{6}{3}\left(x-0\right)+0,\:\quad \:y=-\frac{6}{3}\left(x-0\right)+0\\y=2x,\:\quad \:y=-2x$

b) The length of the transverse axis is given by $2a$ . Therefore, the lenght is 6.

c) See below.

4 0

3 years ago