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2 years ago

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3. For the graph, complete the following: (a) Write a table of ordered pairs from the graph. (b) Write an equation from the tabl

e. (c) State which variable is the independent variable and which variable is the dependent variable. Answer:

1 answer:

alexandr1967 [171]2 years ago

6 0

for this i am working on this um please wait

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THE vertices of a triangle are P(-1,3), Q(2,-1), and R(5,3) choose all the congruent sides of the triangle

Ratling [72]

Answer:

PQ and QR are congruent.

Step-by-step explanation:

The length of PQ = sqrt [(2 - -1)^2 + (-1 - 3)^2]

= sqrt 25

= 5 units.

QR = sqrt [(5-2)^2 + (3 - -1)^2) ]

= sqrt 25

= 5 units.

PR = sqrt [ ( 3-3^2 + (5- -1)^2]

= sqrt 36

= 6 units.

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3 years ago

Write and solve an inequality for the following sentence: Jan has saved $50 in her saving account and needs to save at least $26

Juliette [100K]

50 + x ≥ 268

X ≥ 218

STEP BY STEP WORK

50 + x ≥268
-50 -50

x ≥218

Jan needs to save at least $218 for camp

Hope I helped :)

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3 years ago

Find the HCF and LCM of: x^3+27,2x^3-6x^2+18x,x^2-3x+9

r-ruslan [8.4K]

Answer:

hcf:3x^3-6x^2+y^2-3x+54

lcm:2(x+3)(x^2-3x+9)(x^3-3x^2+9)(y^2-3x+9)

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2 years ago

On a finance exam, 15 accounting majors had an average grade of 90. On the same exam, 7 marketing majors averaged 85, and 10 fin

lilavasa [31]

Answer:

89.84375

Step-by-step explanation:

Given that on a finance exam, 15 accounting majors had an average grade of 90. On the same exam, 7 marketing majors averaged 85, and 10 finance majors averaged 93.

To find weighted mean.

Here weights can be considered as number of majors in each subject

$Marks x 90 85 93\\Weights 15 7 10\\Product 1350 595 930\\$

Weighted total = 1350+595+930 = 2875

Weighted mean = weighted total/total students

$=\frac{2875}{32} =89.84375$

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3 years ago

A teacher randomly selects one student from a group of students. 12 students are wearing a black shirt, 8 are wearing a red shir

Blababa [14]

12+8+8=28
8 students are wearing a blue shirt so the probability is 8/28

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3 years ago

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