All categories

3 years ago

Can anyone give me all answers please hopefully it shows the right thing

Mathematics

1 answer:

sertanlavr [38]3 years ago

5 0

Answer:

It only shows, "Download PDF"

Step-by-step explanation:

Sorry!!! Can't help!

Try taking a pic of it

You might be interested in

What is the prudct of 7\9 and 6

mezya [45]

Answer:

378

Step-by-step explanation:

9 × 7 × 6 = 378

pls mark brainliest, if not, its ok :)

6 0

3 years ago

Read 2 more answers

Round 38.49 to the nearest whole number

AnnyKZ [126]

38. The nearest whole number to 38.49 is 38 because at .5 you begin rounding up to the nearest whole number.

8 0

3 years ago

NEED HELP ASAP PLEASE

ki77a [65]

D.) is the right answer

5 0

3 years ago

Read 2 more answers

Simplify: –3(y 2)2 – 5 6y What is the simplified product in standard form? y2 y.

s344n2d4d5 [400]

The simplified product in standard form is ..

-3y^2 + -6y + -17

Edge 2022!

6 0

2 years ago

Question regarding logarithms.

Eddi Din [679]

$5^{x-2}-7^{x-3}=7^{x-5}+11\cdot5^{x-4}\\\\5^{x-2}-7^{x-2-1}=7^{x-2-3}+11\cdot5^{x-2-2}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\5^{x-2}-\dfrac{7^{x-2}}{7^1}=\dfrac{7^{x-2}}{7^3}+11\cdot\dfrac{5^{x-2}}{5^2}\\\\5^{x-2}-\dfrac{1}{7}\cdot7^{x-2}=\dfrac{1}{343}\cdot7^{x-2}+\dfrac{11}{25}\cdot5^{x-2}\\\\-\dfrac{1}{7}\cdot7^{x-2}-\dfrac{1}{343}\cdot7^{x-2}=\dfrac{11}{25}\cdot5^{x-2}-5^{x-2}\\\\\left(-\dfrac{1}{7}-\dfrac{1}{343}\right)\cdot7^{x-2}=\left(\dfrac{11}{25}-1\right)\cdot5^{x-2}$

$\left(-\dfrac{49}{343}-\dfrac{1}{343}\right)\cdot7^{x-2}=-\dfrac{14}{25}\cdot5^{x-2}\\\\-\dfrac{50}{343}\cdot7^{x-2}=-\dfrac{14}{25}\cdot5^{x-2}\qquad\text{multiply both sides by}\ \left(-\dfrac{25}{14}\right)\\\\\dfrac{50\cdot25}{343\cdot14}\cdot7^{x-2}=5^{x-2}\qquad\text{divide both sides by}\ 7^{x-2}\\\\\dfrac{25\cdot25}{343\cdot7}=\dfrac{5^{x-2}}{7^{x-2}}\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}$

$\dfrac{5^2\cdot5^2}{7^3\cdot7}=\left(\dfrac{5}{7}\right)^{x-2}\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\\dfrac{5^4}{7^4}=\left(\dfrac{5}{7}\right)^{x-2}\\\\\left(\dfrac{5}{7}\right)^4=\left(\dfrac{5}{7}\right)^{x-2}\iff x-2=4\qquad\text{add 2 to both sides}\\\\\boxed{x=6}$

6 0

3 years ago

Read 2 more answers