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3 years ago

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Solve for X. I can't seem to figure out how to isolate the X. The answers of course are X = 0,2 but I need to know the process.

Thank you! $2^x = \frac{3}{2}x + 1$

1 answer:

marin [14]3 years ago

7 0

There is no algebraic way of isolating X unfortunately.
For these equations, you have to solve with numerical approximation.
I suggest Newtons method if you know calculus.

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10m3 of air weigh 13 kg. Chapter Reference c How much does the air weigh in a room 4.2m long, 3.5m wide, and 2.6m high?

Zarrin [17]

Answer:

The air weight in a room is 49.69 kg.

Step-by-step explanation:

It is given that the room is 4.2m long, 3.5m wide, and 2.6m high. It is also given that 10 m³ of air weigh 13 kg.

The formula to find the volume of cuboid is,

$V=l\times b\times h$

$V=4.2\times 3.5\times 2.6$

$V=38.22$

The volume of room is 38.22 m³.

Since it is given that, 10 m³ of air weigh 13 kg. So,

$1 m^3=\frac{13}{10}kg$

$1 m^3=1.3kg$

So the air weight in the room is,

$\text{Air weight}=1.3\times 38.22$

$\text{Air weight}=49.686\approx 49.69$

Therefore the air weight in a room is 49.69 kg.

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3 years ago

A prism is completely filled with 2304 cubes that have edge length of 14 cm What is the volume of the prism?

Tasya [4]

the prism is filled completely, that means volume of prism = sum volume of cubes
volume of a cube =h^3, volume of each cube = 14cmx14cmx14cm=2744cm^3
volume of 2304cubes = 2304 x 2744cm^3=6322176cm^3
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4 years ago

Read 2 more answers

Operations with rational expressions

Scorpion4ik [409]

$$\frac{50-2 w^{2}}{3 w^{2}+9 w-30} \cdot \frac{w^{2}+5 w-14}{6 w-30}=-\frac{w+7}{9}$

Solution:

Given expression:

$$\frac{50-2 w^{2}}{3 w^{2}+9 w-30} \cdot \frac{w^{2}+5 w-14}{6 w-30}$

To solve the given expression:

First simplify: $\frac{50-2 w^{2}}{3 w^{2}+9 w-30}$

$$\frac{50-2 w^{2}}{3 w^{2}+9 w-30}=-\frac{2(w+5)(w-5)}{3(w-2)(w+5)}$

Cancel the common factor (w + 5).

$$=-\frac{2(w-5)}{3(w-2)}$

Now substitute this in the given expression.

$$\frac{50-2 w^{2}}{3 w^{2}+9 w-30} \cdot \frac{w^{2}+5 w-14}{6 w-30}=-\frac{2(w-5)}{3(w-2)} \cdot \frac{w^{2}+5 w-14}{6 w-30}$

Multiply the fractions $\frac{a}{b} \cdot \frac{c}{d}=\frac{a \cdot c}{b \cdot d}$

$$=-\frac{2(w-5)\left(w^{2}+5 w-14\right)}{3(w-2)(6 w-30)}$

Factor the denominator $3(w-2)(6 w-30) =18(w-2)(w-5)$

$$=-\frac{2(w-5)\left(w^{2}+5 w-14\right)}{18(w-2)(w-5)}$

Cancel the common factor 2(w – 5).

$$=-\frac{w^{2}+5 w-14}{9(w-2)}$

Factor the numerator $w^{2}+5 w-14=(w-2)(w+7)$

$$=-\frac{(w-2)(w+7)}{9(w-2)}$

Cancel the common factor (w – 2).

$$=-\frac{w+7}{9}$

$$\frac{50-2 w^{2}}{3 w^{2}+9 w-30} \cdot \frac{w^{2}+5 w-14}{6 w-30}=-\frac{w+7}{9}$

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3 years ago

Factor completely x2 + 8x - 20

kvv77 [185]

X-2
Hope this helps!
Vote me brainliest!

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3 years ago

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3 years ago