All categories

3 years ago

Solve the equation r ÷ (-8) =5

Mathematics

2 answers:

krok68 [10]3 years ago

8 0

Answer:

r=-40

Step-by-step explanation:

r/-8=5

times -8 on both sides

5 times -8 =-40

r=-40

lara [203]3 years ago

7 0

Answer:

r= 13,3

Step-by-step explanation:

let me know if that was wrong good luck

You might be interested in

Se van a unir dos tramos de cadena: uno mide 0.75 m de largo y el otro 15/100 de metro . ¿Cual será la longitud total ya que est

likoan [24]

Step-by-step explanation:

It is given that,

The length of section 1 is 0.75 m and the length of section 2 is 15/100 m. The total length of the section is equal to the sum of section 1 and section 2. So, we will add them as follows :

$l=0.75\ m+\dfrac{15}{100}\ m\\\\\because \dfrac{15}{100}=0.15\\\\l=0.75\ m +0.15\ m\\\\l=0.9\ m$

So, the length of the section is 0.9 m or 9/10 m

8 0

3 years ago

Solve the problem below.

maxonik [38]

Answer:

Please post one by one answer

6 0

2 years ago

Express the given quantity as a single logarithm and simplify: 3logx+2log(y-2)-5logx

storchak [24]

Simplify each term.
Simplify 3log(x) by moving 3 inside the logarithm.
log(x^3)+2log(y−1)−5log(x)

Simplify 2log(y−1) by moving 2 inside the logarithm.
log(x^3)+log((y−1)^2)−5log(x)

Rewrite (y−1)^2 as (y−1)(y−1).
log(x^3)+log((y−1)(y−1))−5log(x)

Expand (y−1)(y−1) using the FOIL Method.
log(x^3)+log(y(y)+y(−1)−1(y)−1(−1))−5log(x)

Simplify each term.
log(x^3)+log(y^2−2y+1)+log(x^−5)

Remove the negative exponent by rewriting x^−5 as 1/x^5.
log(x^3)+log(y^2−2y+1)+log(1/x^5)

Combine logs to get log(x^3(y^2−2y+1))
log(x^3(y^2−2y+1))+log(1/x^5)

Combine logs to get log(x^3(y^2−2y+1)/x^5)
log(x^3(y^2−2y+1)/x^5)

Cancel x^3 in the numerator and denominator.
log(y^2−2y+1/x^2)

Rewrite 1 as 1^2.
y^2−2y+1^2/x^2

Factor by perfect square rule.
(y−1)^2/x^2

Replace into larger expression.

log((y−1)^2/x^2)

3 0

3 years ago

Fill in Sin, Cos, and tan ratio for angle x. Sin X = 4/5 (28/35 simplified)

Fantom [35]

Answer:

Given: $\sin(x) = (4/5)$ .

Assuming that $0 < x < 90^{\circ}$ , $\cos(x) = (3/5)$ while $\tan(x) = (4/3)$ .

Step-by-step explanation:

By the Pythagorean identity $\sin^{2}(x) + \cos^{2}(x) = 1$ .

Assuming that $0 < x < 90^{\circ}$ , $0 < \cos(x) < 1$ .

Rearrange the Pythagorean identity to find an expression for $\cos(x)$ .

$\cos^{2}(x) = 1 - \sin^{2}(x)$ .

Given that $0 < \cos(x) < 1$ :

$\begin{aligned} &\cos(x) \\ &= \sqrt{1 - \sin^{2}(x)} \\ &= \sqrt{1 - \left(\frac{4}{5}\right)^{2}} \\ &= \sqrt{1 - \frac{16}{25}} \\ &= \frac{3}{5}\end{aligned}$ .