$8^{3x-10}=4^{6x-27}\\\\(2^3)^{3x-10}=(2^2)^{6x-27}\qquad\text{use}\ (a^n)^m=a^{nm}\\\\2^{3(3x-10)}=2^{2(6x-27)}\qquad\text{use distributive property}\\\\2^{(3)(3x)+(3)(-10)}=2^{(2)(6x)+(2)(-27)}\\\\2^{9x-30}=2^{12x-54}\iff9x-30=12x-54\qquad\text{add 30 to both sides}\\\\9x=12x-24\qquad\text{subtract 12x from both sides}\\\\-3x=-24\qquad\text{divide both sides by (-3)}\\\\\boxed{x=8}$

6 0

3 years ago

I need help ASAP. The assignment is due at 11:59

Oksanka [162]

Answer:

its answer is 75 i guess

5 0

3 years ago

Read 2 more answers

I need help asap pls and thank you ;)

olga55 [171]

Answer:

$\text{Length of AB is }\frac{ah}{a+h}$

Step-by-step explanation:

Given △KMN, ABCD is a square where KN=a, MP⊥KN, MP=h.

we have to find the length of AB.

Let the side of square i.e AB is x units.

As ADCB is a square ⇒ ∠CDN=90°⇒∠CDP=90°

⇒ CP||MP||AB

In ΔMNP and ΔCND

∠NCD=∠NMP (∵ corresponding angles)

∠NDC=∠NPM (∵ corresponding angles)

By AA similarity rule, ΔMNP~ΔCND

Also, ΔKAP~ΔKPM by similarity rule as above.

Hence, corresponding sides are in proportion

$\frac{ND}{NP}=\frac{CD}{MP} \thinspace\thinspace and\thinspace\thinspace \frac{KA}{KP}=\frac{AB}{PM} \\\\\frac{ND}{NP}=\frac{x}{h} \thinspace\thinspace and\thinspace\thinspace \frac{KA}{KP}=\frac{x}{h}\\\\\frac{NP}{ND}=\frac{h}{x} \thinspace\thinspace and\thinspace\thinspace \frac{KP}{KA}=\frac{h}{x}\\\\\frac{PD}{ND}=\frac{h}{x}-1 \thinspace\thinspace and\thinspace\thinspace \frac{AP}{KA}=\frac{h}{x}-1\\$