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denis23 [38]
3 years ago
11

Walnut High Schools Enrollment is exactly five times as large as the enrollment at walmut junior high the total enrollment for t

he two schools is 852 what is the enrollment at each school​
Mathematics
1 answer:
Vaselesa [24]3 years ago
6 0

Answer:

142

Step-by-step explanation:

If the enrollment is five times as big, then that is six different things. So, take 852 and divide it by 6 to get 142. Or in other words:

852÷6=142

142x6=856

I hope this helps. Cheers^^

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The graph of the distance a car travels at a constant speed passes through the points (0,0) and 2,136). The x scale is hours and
Zepler [3.9K]

We are given that

y-scale is a distance in miles

x-scale is a time in hours

and we are given

The graph of the distance a car travels at a constant speed passes through the points (0,0) and (2,136)

so, points would be

(0,0)

x1=0 , y2=0

(2,136)

x2=2 , y2=136

so, slope would be speed

because speed = distance / time

we can use slope formula

m=\frac{y_2-y_1}{x_2-x_1}

now, we can plug values

m=\frac{136-0}{2-0}

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so, option-B..........Answer


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About Exercise 2.3.1: Proving conditional statements by contrapositive Prove each statement by contrapositive
Sladkaya [172]

Answer:

See proofs below

Step-by-step explanation:

A proof by counterpositive consists on assuming the negation of the conclusion and proving the negation of the hypothesis.

a) Assume that n is not odd. Then n is even, that is, n=2k for some integer k. Hence n²=4k²=2(2k²)=2t for some integer t=2k². Then n² is even, therefore n² is not odd. We have proved the counterpositive of this statement.

b) Assume that n is not even, then n is odd. Thus, n=2k+1 for some integer k. Now, n³=(2k+1)³=8k³+6k²+6k+1=2(4k³+3k²+3k)+1=2t+1 for the integer t=4k³+3k²+3k. Thus n³ is odd, that is, n³ is not even.

c) Suppose that n is not odd, that is, n is even. Now, n=2k for some integer k. Then 5n+3=10k+3=2(5k+1)+1, thus 5n+3 is odd, then 5n+3 is not even.

d) Suppose that n is not odd, then n is even. Now, n=2k for some integer k. Then n²-2n+7=4k²-4k+7=2(2k²-2k+3)+1. Hence n²-2n+7 is odd, that is, n²-2n+7 is not even.

e) Assume that -r is not irrational, then -r is rational. Since -1 is rational, then (-1)(-r)=r is rational. Thus r is not irrational.

f) Assume that 1/z is not irrational. Then 1/z is rational. Multiplucative inverses of rational numbers are rational, hence z is rational, that is, z is not irrational.

g) Suppose that z>y. We will prove that z³+zy²≤z²y+y³ is false, that is, we will prove that z³+zy²>z²y+y³. Multiply by the nonnegative number z² in the inequality z>y to get z³>z²y (here we assume z and y nonzero, in this case either z³>0=y³ is true or z³=0>y³ is true). On the other hand, multiply by z² (positive number) to get zy²>y³. Add both inequalities to obtain z³+zy²>z²y+y³ as required.

h) Suppose than n is even. Then n=2k, and n²=4k² is divisible by 4.

i) Assume that "z is irrational or y is irrational" is false. Then z is rational and y is rational. Rational numbers are closed under sum, then z+y is rational, that is, z+y is not irrational.

3 0
2 years ago
Put your answer in standard notation to 4 decimal places.)
ArbitrLikvidat [17]
3.8275×10−1 :))) there ya go
8 0
3 years ago
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