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3 years ago

Can someone please help me with this i don’t understand:(

Mathematics

1 answer:

swat323 years ago

7 0

Answer:

domain = neg infinity, infinity

range = neg infinity, infinity

Step-by-step explanation:

domain is asking x values which the graph will pass through. since the line is continuous, it touches EVERY x value, same with range and y values.

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PLEASE ANSWER PROPERLY The law of cosines is used to find the measure of Z.

gladu [14]

16² = 18² + 19² - 2(18)(19)cos(z)

256 = 324 + 361 - (684)cos(z)

256 = 685 - (684)cos(z)

-429 = -684cos(z)

cos(z) = -429/-684

z = cos⁻¹ (0.627)

z = cos⁻¹ (cos 51) [ Cos 51 = 0.627 ]

z = 51

In short, Your Answer would be: 51

Hope this helps!

3 0

3 years ago

In the past month, Amy rented 7 video games and 4 DVDs. The rental price for each video game was $3.30. The rental price for eac

Lera25 [3.4K]

Answer:

$39.90

Step-by-step explanation:

if each video game cost $3.30, and she rented 7, then you multiply $3.30 by 7, to get $21.10, and if each DVD was $4.20, and she rented 4, then its $4.20x4= $16.80

after that you just add the two total costs together, which is $23.10+$16.80=$39.90

8 0

3 years ago

Rhett is solving the quadratic equation 0= x2 – 2x – 3 using the quadratic formula. Which shows the correct substitution of the

Brums [2.3K]

A = 1
b = -2
c = -3

And when you plug in and solve you'll get correct answers of x = 3 and x = -1

5 0

3 years ago

Read 2 more answers

What is bigger 5/6 or 13/18?

san4es73 [151]

To compare fractions, you need a common denominator.
6 goes into 18 three times. If we multiply (5/6) by 3 on top and bottom you get
$\frac{15}{18}$
Now we can compare numerators, since 15 > 13
$\frac{5}{6} \ \textgreater \ \frac{13}{18}$

4 0

4 years ago

Simplify your answer as much as possible

loris [4]

9514 1404 393

Answer:

-4u^2/y+(5/2)y^3u^2

Step-by-step explanation:

The applicable rules of exponents are ...

(a^b)(a^c) = a^(b+c)

(a^b)/(a^c) = a^(b-c)

$(8y^2u^7-5y^6u^7)\div(-2y^3u^5)=\dfrac{8y^2u^7-5y^6u^7}{-2y^3u^5}\\\\=-4y^{2-3}u^{7-5}+\dfrac{5}{2}y^{6-3}u^{7-5}=\boxed{\dfrac{-4u^2}{y}+\dfrac{5}{2}y^3u^2}$

7 0

3 years ago