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Mathematics

2 answers:

ivann1987 [24]2 years ago

8 0

Answer: I think it’s a

Step-by-step explanation:

SVETLANKA909090 [29]2 years ago

5 0

Answer: Choice C) $-6 \le x \le 2$

The domain is the set of possible x values or allowed inputs. The left-most point has x = -6 as the smallest allowed x value, while the right-most point has x = 2 as the largest allowed x value.

The domain is the set of numbers between -6 and 2, including both endpoints. So we write $-6 \le x \le 2$

If you want the domain written in interval notation, then you would have [-6, 2]. The square brackets tell the reader "include the endpoint".

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The surface area of the triangular prism is: (see attached image) (in square meters)

Anvisha [2.4K]

The triangle prism should be 60

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3 years ago

I need help on a math question thank u

ElenaW [278]

Answer:

XZ is longer than XY

Step-by-step explanation:

First we can notice all of the answers consist of knowing the lengths, so we must now find the lengths for each segment. To do this, we must use the pythagorean theorem.

(1, -1) and (-5, -5)

-5- -1 = -4

-5-1 = -6

-4^2+-6^2 = c^2

16+36 = c^2

52 = c^2

7.21110255 = c

So, the length of XZ is 7.21110255 units long

(1, -1) and (8, -2)

8-1 = 7

-2--1 = -1

7^2+-1^2 = c^2

49+1 = c^2

50 = c^2

7.07106781 = c

So, the length of XY is 7.07106781 units long.

XZ is longer than XY

4 0

2 years ago

Question regarding logarithms.

Eddi Din [679]

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

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

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