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3 years ago

True or false (if it's false, replace the words in parentheses with the correct answer):

Mathematics

1 answer:

Elis [28]3 years ago

4 0

Answer:

1 is true , 2 is false

Step-by-step explanation:

for equation 2

-2x<8

-2x=8

then we divide both side by coefficient of x which is -2

-2x/-2=8/-2

we will be left with

x= 8/-2

x=-4 which shows that x<-4 and it differs from the second inequality

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I don’t know if the first one is right but could someone guess true or false?

victus00 [196]

Answer:

B) false D) true E) true

Step-by-step explanation:

Plug in numbers to disprove the first one.

Area formula for triangles says D is true

It's the Pythagorean's Theorem for Heaven's sakes.

3 0

4 years ago

What equals 62? and what factors equal 62

nexus9112 [7]

The only factors that equal 62 are

62 • 1 = 62
and
31 • 2 = 62.

Hope this helps!

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4 years ago

Read 2 more answers

I need help in problem 8 plz help me I need a a

iogann1982 [59]

Here are three ways where There could be equal rows

3*12=36
4*9=36
6*6=36

6 0

3 years ago

Anjali paid $8 for a salad. She now has<br> $37. With how much money did she start?

sleet_krkn [62]

Answer:

Below.

Step-by-step explanation:

37+8= 45

45-8=37

so 45.

8 0

3 years ago

Read 2 more answers

HELP TIMER Write the equation of a hyperbola centered at the origin with x-intercept +/- 4 and foci of +/-2(squareroot 5)

meriva

Given:

x-intercepts of the hyperbola are ±4.

The foci of hyperbola are $\pm 2\sqrt{5}$ .

Center of the hyperbola is at origin.

To find:

The equation of the hyperbola.

Solution:

The general equation of a hyperbola:

$\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1$ ...(i)

Where, (h,k) is the center of the hyperbola, ±a are x-intercepts, $(\pm c,0)$ are foci.

Center of the hyperbola is at origin. So, h=0 and k=0.

x-intercepts of the hyperbola are ±4. So,

$\pm a=\pm 4$

$a=4$

The foci of hyperbola are $\pm 2\sqrt{5}$ .

$\pm c=\pm 2\sqrt{5}$

$c=2\sqrt{5}$

We know that,

$a^2+b^2=c^2$

$(4)^2+b^2=(2\sqrt{5})^2$

$16+b^2=20$

$b^2=20-16$

$b^2=4$

Taking square root on both sides, we get

$b=\sqrt{4}$ [b>0]

$b=2$

Substituting $h=0,k=0,a=4,b=2$ in (i), we get

$\dfrac{(x-0)^2}{4^2}-\dfrac{(y-0)^2}{2^2}=1$

$\dfrac{x^2}{4^2}-\dfrac{y^2}{2^2}=1$

Therefore, the correct option is (d).

4 0

3 years ago