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3 years ago

Solve for c C – 7 = -9

Mathematics

2 answers:

ddd [48]3 years ago

8 0

Answer:

$\Huge \boxed{c=-2}$

Step-by-step explanation:

$c-7=-9$

We need to isolate the $c$ variable on one side of the equation.

Adding 7 to both sides of the equation.

$c-7+7=-9+7$

Simplifying the equation.

$c=-2$

Molodets [167]3 years ago

5 0

Answer: -2

Step-by-step explanation:

-2 - 7 = -9

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Solve 1/x-1/x-5=1/3x

Arada [10]

9514 1404 393

Answer:

x = -10

Step-by-step explanation:

We assume your equation is ...

$\dfrac{1}{x}-\dfrac{1}{x-5}=\dfrac{1}{3x}$

Subtracting the right side, then expressing with a common denominator gives ...

$\dfrac{3}{3x}-\dfrac{1}{3x}-\dfrac{1}{x-5}=0\\\\\dfrac{2}{3x}-\dfrac{1}{x-5}=0\\\\\dfrac{2(x-5)-3x}{3x(x-5)}=0\\\\\dfrac{-x-10}{3x(x-5)}=0\\\\\boxed{x=-10} \qquad\text{equate the numerator to 0 and add $x$}$

3 0

2 years ago

Please graph y=2x-5. find all zeros and list one.

mixas84 [53]

The zero of the function y = 2x - 5 is x = 5

<h3>What are linear equations?</h3>

Linear equations are equations that have constant average rates of change. Note that the constant average rates of change can also be regarded as the slope or the gradient

<h3 /><h3>How to determine the solution to the system?</h3>

A system of linear equations is a collection of at least two linear equations.

In this case, the equation is given as

y = 2x - 5

Next, we plot the above equation.

From the graph of the equation, we can see that the graph crosses the y-axis at 5

This represents the zero of the function

Hence, the zero of the function y = 2x - 5 is x = 5

Read more about linear equations at:

brainly.com/question/14323743

#SPJ1

4 0

1 year ago

Determine whether the data set represents a direct variation, an inverse variation, or neither?

arsen [322]

Answer:

suggestions use math way or algebra calculator

5 0

3 years ago

How would i solve this the shaded region is the whole inside circle.

rosijanka [135]

Answer:

34.48cm²

Step-by-step explanation:

Assuming the shaded area doesn't contain the triangle:

area of triangle = bh/2

area of triangle = 4(2)/2

area = 4

area of circle = πr²

area = π3.5²

area = 38.48

area of shaded = 38.48 - 4

area = 34.48cm²

6 0

1 year ago

I really need help with this! Thank you!!

photoshop1234 [79]

<h3>Answer:</h3>

x = 13; NP = 2 18/23 ≈ 2.8; NL = 5 5/23 ≈ 5.2
D. x = 17, y = 5

<h3>Step-by-step explanation:</h3>

1. In order for this problem to be workable, we need to assume PQ ║ ML. Then ΔPNQ ~ ΔLNM and ∠L = ∠P = 60°.

... ∠L = 60°

... (3x+21)° = 60°

... 3x = 39 . . . . . . . divide by °, subtract 21

... x = 13 . . . . . . . . .divide by 3

The side lengths of similar triangles are proportional, so we have ...

... (3.2 cm)/y = (6 cm)/(8-y)

Multiplying by the product of the denominators, and dividing by (cm), we have ...

... 3.2(8 -y) = 6y

... 3.2×8 = 9.2y . . . . . add 3.2y

... y = 3.2×8/9.2 = 64/23 = 2 18/23 ≈ 2.78 = NP

Then ...

... 8-y = NL ≈ 5.22

In summary: x = 13; NP ≈ 2.8; NL ≈ 5.2

_____

2. If we assume figures ABCD and PSRQ are similar, we can find the values of the variables. We assume ∠C ≅ ∠R, so ...

... 4x +27° = 95°

... 4x = 68° . . . . . subtract 27°

... x = 17° . . . . . . . divide by 4

___

AB/AD = PS/PQ . . . . . corresponding sides of similar figures are proportional

... 4y/(3y-5) = 10/5 . . . . . units of ft cancel

... 5×4y = 10(3y -5) . . . . . multiply by 5(3y-5)

... 20y = 30y -50 . . . . . . simplify

... 50 = 10y . . . . . . . . . . . add 50-20y

... 5 = y . . . . . . . . . . . . . . divide by 10

Using this value of y, we have ...

AB = 4y ft = 20 ft; AD = (3y-5) ft = 10 ft. Both these values are double the corresponding lengths on PSRQ.

In summary, x = 17°, y = 5. . . . . . (note that the ° symbol is appropriate for x)

_____

You should have your teacher show you how to work these problems <em>using only the given information</em>.

In the second problem, you cannot start with the assumption that the figures are similar, as that is what you're being asked to prove. Please note, too, that two sides and one angle of a quadrilateral are insufficient to show similarity.

7 0

3 years ago