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3 years ago

A trucking company delivered 4,604 cubic yards of

Mathematics

1 answer:

nadezda [96]3 years ago

6 0

Answer:
13,812 cubic yards of sand

step-by-step explanation:
4,604 cubic yards of gravel
4,604 • 3
13,812

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The ratio of boys to girls in a group is 7:1. If there are 66 more boys than girls, work out how many girls there are

nevsk [136]

Answer:11

Step-by-step explanation:

Boys : girls=7:1

Sum of ratio=7+1=8

Let them number of girls be y

Then they number of boys=y+66

Total number of pupils=y+y+66

Total number of pupils=2y+66

Number of girls=(girls ratio)/(sum of ratio) x (total number of pupils)

y=1/8 x (2y+66)

Cross multiply

y x 8=2y + 66

8y=2y + 66

Collect like terms

8y-2y=66

6y=66

Divide both sides by 6

6y/6=66/6

y=11

The number of girls is 11

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3 years ago

The equations y = tan^2 (x) and y = 1 are graphed below. According to the graphwhich of the following is part the solution to th

SVETLANKA909090 [29]

Answer: a

Step-by-step explanation:

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8 0

3 years ago

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A coffe table is on sale for 20% off the orgingial price of $115. If there is 7.5% sales tax, what is the original price of the

Anuta_ua [19.1K]

Answer: 98.90

Step-by-step explanation:

115*.2=23

115-23=92

0.075*92=6.9

6.90+92=$98.90

3 0

3 years ago

How would you find the parallel and perpendicular line of y=4x+3

kirza4 [7]

Parallel would be anything with the slope of 4x and perpendicular would be anything with the slope of 1/4

5 0

3 years ago

Given that 'n' is a natural number. Prove that the equation below is true using mathematical induction.

LenaWriter [7]

<h3>To ProvE :- </h3>

1 + 3 + 5 + ..... + (2n - 1) = n²

Method :-

If P(n) is a statement such that ,

P(n) is true for n = 1
P(n) is true for n = k + 1 , when it's true for n = k ( k is a natural number ) , then the statement is true for all natural numbers .

$\sf\to \textsf{ Let P(n) : 1 + 3 + 5 + $\dots$ +(2n-1) = n$^{\sf 2}$ }$

Step 1 : Put n = 1 :-

$\sf\longrightarrow LHS = \boxed{\sf 1 } \\$

$\sf\longrightarrow RHS = n^2 = 1^2 = \boxed{\sf 1 }$

Step 2 : Assume that P(n) is true for n = k :-

$\sf\longrightarrow 1 + 3 + 5 + \dots + (2k - 1 ) = k^2$

Add (2k +1) to both sides .

$\sf\longrightarrow 1 + 3+5+\dots+(2k-1)+(2k+1)=k^2+(2k+1)$

RHS is in the form of ( a + b)² = a²+b²+2ab .

$\sf\longrightarrow 1 + 3+5+\dots+(2k-1)+(2k+1)= (k +1)^2$

Adding and subtracting 1 to LHS .

$\sf\longrightarrow 1 + 3+5+\dots+(2k-1)+(2k+1) + 1 -1 = (k +1)^2 \\$

$\sf\longrightarrow 1 + 3+5+\dots+(2k-1)+(2k+2) - 1 = (k +1)^2$

Take out 2 as common .

$\sf\longrightarrow 1 + 3+5+\dots+(2k-1)+\{2(k+1)-1\}= (k +1)^2$

P(n) is true for n = k + 1 .

Hence by the principal of Mathematical Induction we can say that P(n) is true for all natural numbers 'n' .

**Edits are welcomed**

8 0

2 years ago

Read 2 more answers