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2 years ago

Show your work please no links please!:)

Mathematics

1 answer:

BabaBlast [244]2 years ago

8 0

Answer:

hgoahdjfgggfhhfjfhfjf mark me brainliestplz

Step-by-step explanation:

You might be interested in

Use the distance formula to decide whether ab=bc <br> A(4,2)<br> B(-1,-1)<br> C(-6,-4)

quester [9]

$AB= \sqrt{(-1-4)^2+(-1-2) ^{2} }= \sqrt{(-5)^2+(-3)^2}= \sqrt{34} \\ \\ BC= \sqrt{(-6+1)^2+(-4+1) ^{2} }= \sqrt{(-5)^2+(-3)^2}= \sqrt{34} \\ \\ \sqrt{34}= \sqrt{34} \ \ \to \ \ AB=BC$

4 0

2 years ago

F(x) = -5x +8(7-x) find x such that f(x) =160

s344n2d4d5 [400]

Answer:

Step-by-step explanation:

f(x) = -5x + 8(7 - x)......when f(x) = 160

160 = -5x + 8(7 - x)

160 = -5x + 56 - 8x

160 - 56 = -5x - 8x

104 = - 13x

-104/13 = x

-8 = x <====

7 0

2 years ago

What is the following sum? Assume x greater-than-or-equal-to 0 and y greater-than-or-equal-to 0

Free_Kalibri [48]

Answer:

$\sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}=2xy^2\sqrt{x}+2xy\sqrt{y}$

Hence, ption B is true.

Step-by-step explanation:

Given the expression

$\sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}$

solving the expression

$\sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}$

$\sqrt{x^2y^3}=xy\sqrt{y}$

$2\sqrt{x^3y^4}=2xy^2\sqrt{x}$

so the expression becomes

$\:\sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}=xy\sqrt{y}+2xy^2\sqrt{x}+xy\sqrt{y}$

Group like terms

$=2xy^2\sqrt{x}+xy\sqrt{y}+xy\sqrt{y}$

Add similar elements

$=2xy^2\sqrt{x}+2xy\sqrt{y}$

Therefore, we conclude that:

$\sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}=2xy^2\sqrt{x}+2xy\sqrt{y}$

Hence, option B is true.

6 0

2 years ago

Read 2 more answers

A professor records the difference between the marks scored by the students in the class test last week and those scored in the

dezoksy [38]

Answer: $(10.82\ , 13.18)$

Step-by-step explanation:

The confidence interval for population mean is given by :-

$\mu\pm\ z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

Given : Sample size : n= 35 , large sample (n>30)

Mean difference : $\overline{x}=12$

Standard deviation : $\sigma=3$

Significance level : $\alpha=- 1-0.999=0.001$

Critical value : $z_{\alpha/2}=z_{0.0005}=3.4807567$

Now, the 99.9% confidence interval for the mean difference between the marks scored last week and marks scored this week by all the students will be :-

$12\pm\ (3.4807567)\dfrac{2}{\sqrt{35}}\\\\=12\pm1.18\\\\=(10.82\ , 13.18)$

Hence, the 99.9% confidence interval for the mean difference between the marks scored last week and marks scored this week by all the students = $(10.82\ , 13.18)$

4 0

2 years ago

in triangle ABC shown below, L is the midpoint of BC, M is the midpoint of AB, and N is the midpoint of AC. Find the perimeter o

alexdok [17]

since m and n are the midpoints of ab and ac resistively

then ,bc=2mn=16

since m and l are the midpoints of ba and bc respectively

then, ac =2mn =10

then,nc =1/2 × 10 =5

similary, mb=6

then ,perimeter=nc+mb+mn+bc=5+6+8+16=35

6 0

2 years ago