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3 years ago

How many different ways can you represent the weight in 1000g?

Mathematics

1 answer:

Sonbull [250]3 years ago

3 0

8 different ways because there are 8 units for measuring weight.

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A) Find the distance of line segment PR. Round answer to the nearest tenth. B) Find the distance around the park (perimeter).

juin [17]

Answer:

A. $PR = 80.6 yd$

B. Perimeter = 190.6 yd

Step-by-step explanation:

A. Distance between P(10, 50) and R(80, 10):

$PR = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$PR = \sqrt{(80 - 10)^2 + (10 - 50)^2}$

$PR = \sqrt{(70)^2 + (-40)^2}$

$PR = \sqrt{4,900 + 1,600}$

$PR = \sqrt{6,500}$

$PR = 80.6 yd$ (nearest tenth)

B. Distance around the park = Perimeter = PR + PQ + QR

PR = 80.6 yd

PQ = |50 - 10| = 40 yd

QR = |80 - 10| = 70 yd

Perimeter = 80.6 + 40 + 70

Perimeter = 190.6 yd

7 0

3 years ago

in your economics 101 class you have scores of 68,82,87 and 89 on the first four of five tests to earn a grade of a B or higher

Nadusha1986 [10]

(68 + 82 + 87 + 89 + x) / 5 > = 80
(326 + x) / 5 > = 80
326 + x > = 80 * 5
326 + x > = 400
x > = 400 - 326
x > = 74.......u would have to make at least a 74 to earn a B

3 0

3 years ago

the sum of 2 consecutive integers is at most the difference between nine times the smaller and 5 times the larger

Reil [10]

<h2>Answer:</h2>

$x\geq 3 \ and \ x+1 \geq 4$

<h2>Step-by-step explanation:</h2>

The question in this problem is:

<em>The sum of 2 consecutive integers is at most the difference between nine times the smaller and 5 times the larger. What are the numbers?</em>

<em />

First of all, let's name the first variable $x$ which is the smaller number. Accordingly, the lager number will be $x+1$ given that those numbers are consecutive. On the other hand<em> at most </em>conveys the idea of an inequality, which is:

$\leq \\ which \ means \ less \ than$

So:

1. The sum of 2 consecutive integers can be written as:

$v+(v+1)$

2. Nine times the smaller and 5 times the larger can be written as:

$9v-5(v+1)$

Finally, the whole statement:

The sum of 2 consecutive integers is at most the difference between nine times the smaller and 5 times the larger:

$x+(x+1) \leq 9x-5(x+1) \\ \\ x+x+1\leq 9x-5x-5 \\ \\ 2x+1 \leq 4x-5 \\ \\ 6 \leq 2x \\ \\$

$x+(x+1) \leq 9x-5(x+1) \\ \\ x+x+1\leq 9x-5x-5 \\ \\ 2x+1 \leq 4x-5 \\ \\ 6 \leq 2x \\ \\ \frac{6}{2} \leq \frac{2x}{2} \\ \\ 3 \leq x \\ \\ x\geq 3 \\ \\ and \\ \\ x+1 \geq 4$

The two numbers are:

$x\geq 3 \ and \ x+1 \geq 4$

6 0

4 years ago

Help me please solve the following equation for x.

wlad13 [49]

Answer:

x = 13

x = -5

Step-by-step explanation:

$\frac{(x-4)^2}{3}-13=14\\\frac{(x-4)^2}{3}=27\\(x-4)^2=81\\x-4=\sqrt{81}\\\\ x-4=9\\x=13\\\\x-4=-9\\x=-5$

5 0

3 years ago

Math problem help please

Harman [31]

Answer:

One ordered pair would be (1, -6).

Step-by-step explanation:

Let x = 1, then

3(1) - y = 9

-y = 9 - 3 = 6

y = -6.

5 0

3 years ago

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How many different ways can you represent the weight in 1000g?​

How many different ways can you represent the weight in 1000g?