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3 years ago

X-(3-2x)-(4-5x)=-7. what is the value of x?

Mathematics

1 answer:

pshichka [43]3 years ago

4 0

Answer:

x=0

Step-by-step explanation:

expand x-(3-2x)-(4-5x)=-7

x-3+2x-4+5x=-7

8x-7=-7

8x-7+7=0

8x=0

x=0

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RSTUV - EABCD. Find AE.

larisa86 [58]

9514 1404 393

Answer:

AE = 3

Step-by-step explanation:

First of all, we need to find corresponding sides that are defined in both figures. The table below shows the given values.

From the table, we can write the proportion ...

EA/RS = BC/TU . . . corresponding sides are proportional

EA/6 = 4/8 . . . . . . . substitute given lengths

EA = 6(4/8) = 3

The length of AE is 3 units.

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2 years ago

A game room has a floor that is 120 feet by 75 feet.A scale drawing of the floor on grid paper uses a scale of 1 unit:5 feet.Wha

erik [133]

We know,
1 unit : 5 feet

so, 120 ft = (120 / 5) units
= 24 units

75 ft = (75 / 5) units
= 15 units

∴ Dimensions of scale drawing is 24 units by 15 units.

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3 years ago

How would the expression x^3-8 be written using difference of cubes

ludmilkaskok [199]

we can use formula

$a^3-b^3=(a-b)(a^2+ab+b^2)$

we can compare

we get

a=x

b=2

now, we can plug that in formula

and we get

$x^3-2^3=(x-2)(x^2+2x+2^2)$

now, we can simplify it

$x^3-8=(x-2)(x^2+2x+4)$ ...........Answer

4 0

3 years ago

Read 2 more answers

Simplify the variable expression by evaluating its numerical part, and enter

Ksju [112]

Answer:7w

Step-by-step explanation:

the process is simple 13 minus 6 equals 7 which only leaves w therefore the answers 7w

8 0

3 years ago

Read 2 more answers

he volume of a fish tank is 50 cubic feet. If the density is 0.2 fish over feet cubed, how many fish are in the tank?

V125BC [204]

There are 10 fish in the tank

Solution:

Given that, volume of a tank is 50 cubic feet

Density is 0.2 fish over feet cubed

To find: Number of fish in the tank

Number of fish in the tank is found by multiplying the volume of tank and density

The formula is given by:

$\text{Number of fish} = \tex{\text{volume of tank}} \times {\text{density of fish over feet cubed}}$

Substituting the given values, we get

$\text{Number of fish} = 50 ft^3 \times \frac{\text{ 0.2 fish}}{ft^3}\\\\ \text{Number of fish} = 50 \times 0.2 \text{ fish }\\\\ \text{Number of fish} = 10 \text{ fish }$

Thus there are 10 fish in the tank

5 0

3 years ago