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madreJ [45]
3 years ago
15

The difference between a number divided by 3 and 4 is 21

Mathematics
1 answer:
Alona [7]3 years ago
8 0

Answer:

3*4=12

12+12=24

24-3=21

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Find three solutions of the equation y=9x-4?
Bas_tet [7]
<span>answer
A) (-5,-49),(-2,-22),(3,23)

</span>when x = - 5 ; y=9x-4 = 9(-5) -4  = -49
when x = - 2 ; y=9x-4 = 9(-2) -4  = -22
when x = 3 ; y=9x-4 = 9(3) -4  = 23
6 0
3 years ago
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A metalworker has a metal alloy that is 15​% copper and another alloy that is 60​% copper. How many kilograms of each alloy shou
coldgirl [10]

18 kg of 15% copper and 72 kg of 60% copper should be combined by the metalworker to create 90 kg of 51% copper alloy.

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

Let x = kg of 15% copper alloy

Let y = kg of 60% copper alloy

Since we need to create 90 kg of alloy we know:

x + y = 90

51% of 90 kg = 45.9 kg of copper

So we're interested in creating 45.9 kg of copper

We need some amount of 15% copper and some amount of 60% copper to create 45.9 kg of copper:

0.15x + 0.60y = 45.9

but

x + y = 90

x= 90 - y

substituting that value in for x

0.15(90 - y) + 0.60y = 45.9

13.5 - 0.15y + 0.60y = 45.9

0.45y = 32.4

y = 72

Substituting this y value to solve for x gives:

x + y = 90

x= 90-72

x=18

Therefore, in order to create 90kg of 51% alloy, we'd need 18 kg of 15% copper and 72 kg of 60% copper.

6 0
3 years ago
Please help me if you don't mind​
yarga [219]
Miya los cento 5-63 9-64
7 0
2 years ago
Suppose that a student has $150 in a band savings account at the start of the school year. Calculate the change in that savings
Roman55 [17]
150 12 x 10=120+150= 270
4 0
3 years ago
Plzzz help plzzzzzzzzzzzzzz
mezya [45]

Given:

The figure of two quadrilaterals.

In ABCD,AB=18,BC=20,CD=22,AD=24

In EFGH,EF=27,FG=30,GH=34, EH=36

To find:

Whether the figures are congruent, similar or neither.

Solution:

Ratio of corresponding sides are:

\dfrac{AB}{EF}=\dfrac{18}{27}

\dfrac{AB}{EF}=\dfrac{2}{3}

Similarly,

\dfrac{BC}{FG}=\dfrac{20}{30}

\dfrac{BC}{FG}=\dfrac{2}{3}

\dfrac{CD}{GH}=\dfrac{22}{34}

\dfrac{CD}{GH}=\dfrac{11}{17}

And,

\dfrac{AD}{EH}=\dfrac{24}{36}

\dfrac{AD}{EH}=\dfrac{2}{3}

Clearly, \dfrac{AB}{EF}=\dfrac{BC}{FG}=\dfrac{AD}{EH}\neq \dfrac{CD}{GH}.

All corresponding sides are not proportional.

Therefore, the figures are neither similar nor congruent. Hence, third option is correct.

7 0
3 years ago
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