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Misha Larkins [42]
3 years ago
5

Find the solution of y = –x – 3 for x = 1.

Mathematics
1 answer:
nikklg [1K]3 years ago
5 0

Answer:

A.  :)

Step-by-step explanation:

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What is the percent decimal and fraction of 12 out of 15
melomori [17]
Percent: 80%

Fraction: 12/15

Decimal: 0.8.    hope that helped.
5 0
4 years ago
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The Bungling Brothers Circus is in town and you are part of the crew that
Kipish [7]

Answer:

a. 80.83 ft b. 40.42 ft

Step-by-step explanation:

Let h = height of pole = 70 ft, L = length of cable and x = distance of cable on ground to pole and Ф = angle between cable and ground.

a) How long is the cable?

Since L, h and x form a right-angled triangle with angle Ф and h being the opposite side to Ф and L being the hypotenuse side, by trigonometric ratios,

sinФ = h/L

L = h/sinФ

L = 70 ft/sin60°

L = 70 ft/0.8660

L = 80.83 ft

b) How far from the pole should the cable be attached to the ground?

Since L, h and x form a right-angled triangle with angle Ф and h being the opposite side to Ф and x being the adjacent side, by trigonometric ratios,

tanФ = h/x

x = h/tanФ

x = 70 ft/tan60°

x = 70 ft/1.7321

x = 40.42 ft

8 0
3 years ago
1 2 3 4 5 6 7 8 9 10
swat32

Using the conversion of exponent to power, the equivalent expression is given as follows:

\frac{\sqrt{x}}{\sqrt[4]{y}z^2}

<h3>How to transform an exponent to a power?</h3>

It happens according to the following rule, with the denominator as the root and the numerator as the exponent:

a^{\frac{n}{m}} = \sqrt[m]{a^n}

In this problem, we are given the following expression:

x^{\frac{1}{2}}y^{-\frac{1}{4}}z^{-2}

Negative exponents go to the denominator, hence the <em>equivalent expression</em> is given by:

\frac{\sqrt{x}}{\sqrt[4]{y}z^2}

More can be learned about the conversion of exponent to power at brainly.com/question/2020414

#SPJ1

7 0
2 years ago
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Someone please help 10 points each Please help me outttt
creativ13 [48]

Answer:

a, 1,<Z b, 1,zy c,XZY

2,<Y 2, zx

3, <X 3, xy

8 0
3 years ago
What are the domain range and asymptote of h(x)=6^x-4
Maru [420]

The asymptote of the function f(x) = aˣ is y = 0.

The domain of the function f(x) = aˣ is x ∈ R.

The range of the function f(x) = aˣ is y > 0.

-----------------------------------------------------------------------------------

f(x - n) - shifting the graph by n units to the right

f(x + n) - shifting the graph by n units to the left

f(x) - n - shifting the graph by n units down

f(x) + n - shifting the graph by n units up

---------------------------------------------------------------------------

We have h(x)=6^{x-4}

f(x)=6^x\to f(x-4)=6^{x-4} - shifting the graph of f(x) = 6ˣ, 4 units to the right. Therefore

Domain - no change

Range - no change

Asymptote - no change

Answer:

The asymptote is y = 0. The domain is x ∈ R. The range is y > 0.

If h(x)=6^x-4, therefore

f(x)=6^x\to f(x)-4=6^x-4 - shifting the graph of f(x) = 6ˣ, 4 units down.

Therefore, yor answer is:

Domain - no change (x ∈ R)

Range - change → y > -4

Asymptote - change → y = -4


3 0
4 years ago
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