All categories

3 years ago

PLS HELP.

Mathematics

2 answers:

djyliett [7]3 years ago

8 0

Answer:

y=1/10 times 5

Step-by-step explanation:

you would times by 5 on both sides

Alexandra [31]3 years ago

6 0

Step-by-step explanation:

y/-5 = 1/10

10y = 1*-5

y = - 5/10

y = - 1/2

You might be interested in

Write an equality that represents the phrase "the sum of 4 and x is greater than 6." Draw a graph that represents the inequality

Romashka [77]

Answer:

(4+x)>6

See "the sum of 4 and x" means '4+x' and "greater than 6" means the other number ,aka x which equals 4+x, is greater than six. So, the answer comes out to be (4+x)>6.

3 0

3 years ago

What is four and one over five divided bye one and two overfive

marshall27 [118]

Can you type it out with actual numbers instead of letters please? Thank you

7 0

3 years ago

Plsssssssssssss helpppppppp

STatiana [176]

Answer:

the last one

Step-by-step explanation:

5 0

3 years ago

There are 9 yellow and 4 purple balls in a bag. If you randomly choose balls one at a time, with replacement, what is the probab

ahrayia [7]

Did you get a answer?

7 0

3 years ago

Read 2 more answers

Write the equation of a hyperbola centered at the origin with x-intercepts +/- 4 and foci of +/-2(sqrt5)

Xelga [282]

Answer:

$\frac{x^2}{16}-\frac{b^2}{4}=1$

Step-by-step explanation:

A hyperbola is the locus of a point such that its distance from a point to two points (known as foci) is a positive constant.

The standard equation of a hyperbola centered at the origin with transverse on the x axis is given as:

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

The coordinates of the foci is at (±c, 0), where c² = a² + b²

Given that a hyperbola centered at the origin with x-intercepts +/- 4 and foci of +/-2√5. Since the x intercept is ±4, this means that at y = 0, x = 4. Substituting in the standard equation:

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\\\frac{4^2}{a^2}-\frac{0}{b^2} =1\\\frac{4^2}{a^2}=1\\ a^2=16\\a=\sqrt{16}=4\\ a=4$

The foci c is at +/-2√5, using c² = a² + b²:

$c^2=a^2+b^2\\(2\sqrt{5} )^2=4^2+b^2\\20 = 16 + b^2\\b^2=20-16\\b^2=4\\b=\sqrt{4}=2\\ b=2$

Substituting the value of a and b to get the equation of the hyperbola:

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\\\\\frac{x^2}{16}-\frac{b^2}{4}=1$

5 0

3 years ago