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2 years ago

Simplify the expression. c^-6 d^4/c^-16 d^14

Mathematics

1 answer:

pochemuha2 years ago

6 0

Answer:

$\red{\frac{ {c}^{10}}{{d}^{10} } }$

Step-by-step explanation:

$\huge \: \frac{ {c}^{ - 6} {d}^{4}}{ {c}^{ - 16} {d}^{14} } \\ \\ = \huge \: \frac{ {c}^{16 - 6}}{{d}^{14 - 4} } \\ \\ = \huge \: \red{\frac{ {c}^{10}}{{d}^{10} } }$

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Can someone help me with this one question?

kondor19780726 [428]

Answer:

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Step-by-step explanation:

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2 years ago

I) Construct a triangle PQR such that |PQ|=8cm,{RPQ=90°{PQR=30°.Measure |RQ|

scoray [572]

Answer:

6.93 cm

Step-by-step explanation:

You have a right triangle (90°), so you do as follow:

If I understand correctly, you are looking for the hypotenuse so

$cos(30) = \frac{PQ}{QR} = \frac{8 cm}{QR}$

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2 years ago

What does h(40)=1820 mean it terms of problem?!!! HELPP

mylen [45]

Answer:

D is the correct option.

Step-by-step explanation:

In the given table, the first column is for "hours of training" and the second column is for "Monthly pay".

Now, we have to state the meaning of the term h(40)=1820.

In coordinate system we can write this as (40,1820).

Since, the first column represents "hours of training" and second column is for "Monthly pay". Hence, we can say that for 40 hours of training the monthly payment is $1820.

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Therefore, D is the correct option.

5 0

3 years ago

40 Points please help

insens350 [35]

Answer:

Slope = ⅓ Y-intercept = -2

Step-by-step explanation:

by transforming the equation to y= ⅓x -2

then it becomes Y= ⅓x +(-2)

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5 0

2 years ago

Read 2 more answers

If a = √3-√11 and b = 1 /a, then find a² - b²

Serga [27]

If $b=\frac1a$ , then by rationalizing the denominator we can rewrite

$b = \dfrac1{\sqrt3-\sqrt{11}} \times \dfrac{\sqrt3+\sqrt{11}}{\sqrt3+\sqrt{11}} = \dfrac{\sqrt3+\sqrt{11}}{\left(\sqrt3\right)^2-\left(\sqrt{11}\right)^2} = -\dfrac{\sqrt3+\sqrt{11}}8$

Now,

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

and

$a - b = \sqrt3 - \sqrt{11} + \dfrac{\sqrt3 + \sqrt{11}}8 = \dfrac{9\sqrt3 - 7\sqrt{11}}8$

$a + b = \sqrt3 - \sqrt{11} - \dfrac{\sqrt3 + \sqrt{11}}8 = \dfrac{7\sqrt3 - 9\sqrt{11}}8$

$\implies a^2 - b^2 = \dfrac{\left(9\sqrt3 - 7\sqrt{11}\right) \left(7\sqrt3 - 9\sqrt{11}\right)}{64} = \boxed{\dfrac{441 - 65\sqrt{33}}{32}}$

5 0

1 year ago