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3 years ago

I need help solving this problem: -1/7c=21. Please help me and solve for c

Mathematics

1 answer:

kvv77 [185]3 years ago

3 0

So you need to get C by itself. So you are going to divide -1/7 by 21. Then that will give you the answer x=3

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each morning, danai buys breakfast on her why to work. In the past thirty days, she bought a bagel on 6 days, a banana on 12 day

Ksju [112]

Answer:

0.7 or 70%

Step-by-step explanation:

Total number of days

6 + 12 + 3 + 9 = 30

The number of days she bought a banana or orange

12 + 9 = 21

The probability of buying a banana or an orange is

P(b or o) = 21/30 = 0.7 = 70%

8 0

2 years ago

If 4 + x = −3 + 2x, then 9x =_____?

gayaneshka [121]

Answer:

4+x=-3+2x

4+3=2x-x

7=x

9x=9*7

=63

Step-by-step explanation:

8 0

3 years ago

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A cable hangs between two poles of equal height and 35 feet apart. At a point on the ground directly under the cable and x feet

gayaneshka [121]

Answer:

293.38 pounds

Step-by-step explanation:

We are given that

Distance between poles=35 feet

$h(x)=10+0.1(x^{1.5})$

Weight of cable=10.4 per linear foot

We have to find the weight of the cable.

Differentiate w.r.t

$h'(x)=0.1(1.5)x^{0.5}=0.15x^{0.5}$

$s=2\int_{0}^{17.5}\sqrt{1+(h'(x))^2}dx$

$s=2\int_{0}^{17.5}\sqrt{1+(0.15x^{0.5})^2}dx$

$s=2\int_{0}^{17.5}\sqrt{1+0.0225x}dx$

Let $1+0.0225x=t$

$dx=\frac{1}{0.0225}dt$

$s=\frac{2}{0.0225}\int_{0}^{17.5}\sqrt{t}dt$

$s=\frac{2}{0.0225}\times\frac{2}{3}[t^{\frac{3}{2}}]^{17.5}_{0}$

$s=2\times \frac{2}{3\times0.0225}[(1+0.0255x)^{\frac{3}{2}]^{17.5}_{0}$

$s=\frac{4}{3\times 0.0225}((1+0.0225(17.5))^{\frac{3}{2}-1)$

s=28.21

Weight of cable= $28.21\times 10.4=293.38$ pound

8 0

3 years ago

Which graph represents the function f (x) = StartFraction 2 Over x minus 1 EndFraction + 4?

Pepsi [2]

Answer:

On a coordinate plane, a hyperbola is shown. One curve opens up and to the right in quadrant 1, and the other curve opens down and to the left in quadrant 3. A vertical asymptote is at x = 1, and the horizontal asymptote is at y = 4

Step-by-step explanation:

The given function is presented as follows;

$f(x) = \dfrac{2}{x - 1} + 4$

From the given function, we have;

When x = 1, the denominator of the fraction, $\dfrac{2}{x - 1}$ , which is (x - 1) = 0, and the function becomes, $\dfrac{2}{1 - 1} + 4 = \dfrac{2}{0} + 4 = \infty + 4 = \infty$ therefore, the function in undefined at x = 1, and the line x = 1 is a vertical asymptote

Also we have that in the given function, as x increases, the fraction $\dfrac{2}{x - 1}$ tends to 0, therefore as x increases, we have;

$\lim_ {x \to \infty} \dfrac{2}{(x - 1)} \to 0, and \ \dfrac{2}{(x - 1)} + 4 \to 4$

Therefore, as x increases, f(x) → 4, and 4 is a horizontal asymptote of the function, forming a curve that opens up and to the right in quadrant 1

When -∞ < x < 1, we also have that as x becomes more negative, f(x) → 4. When x = 0, $\dfrac{2}{0 - 1} + 4 = 2$ . When x approaches 1 from the left, f(x) tends to -∞, forming a curve that opens down and to the left

Therefore, the correct option is on a coordinate plane, a hyperbola is shown. One curve opens up and to the right in quadrant 1, and the other curve opens down and to the left in quadrant 3. A vertical asymptote is at x = 1, and the horizontal asymptote is at y = 4.

5 0

3 years ago

Help me pls I need this grade

Ivahew [28]

yes lol bigger hah trigger lol susush

5 0

3 years ago

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