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Strike441 [17]
3 years ago
13

Choose the correct response​

Mathematics
1 answer:
NeTakaya3 years ago
6 0

Answer:  0

To find this answer, we replace x with 0 in the g(x) function

g(x) = -2x

g(0) = -2*0

g(0) = 0

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PLEASE HELP ASAP
marysya [2.9K]

Answer:

A’ (-5,6) B’ (-5,8) C’ (-1,8) D’ (-1,6)

Step-by-step explanation:


7 0
3 years ago
How many miles do you have to walk in a three day period to reach 6 2/3 miles
hichkok12 [17]

Answer:

approx. 2.2 miles

Step-by-step explanation:

If we put 2/3 into a decimal, (.66) we then add it to the 6 miles, aprox. 6.6 miles. Divide 6.6 by 3,

and you're answer is 2.2.

Approx. 2.2 miles a day over a 3 day period.

4 0
3 years ago
Prove :
Sauron [17]

Answer:

See Below.

Step-by-step explanation:

We want to verify the equation:

\displaystyle \frac{1}{\sec\alpha+1}-\frac{\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha }{\sin^2\alpha }-\frac{1}{\sec\alpha -1}

We can convert sec(α) to 1 / cos(α):

\displaystyle \frac{1}{1/\cos\alpha+1}-\frac{\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha }{\sin^2\alpha }-\frac{1}{\sec\alpha -1}

Multiply both layers of the first fraction by cos(α):

\displaystyle \frac{\cos\alpha}{1+\cos\alpha}-\frac{\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha }{\sin^2\alpha }-\frac{1}{\sec\alpha -1}

Create a common denominator. We can multiply the first fraction by (1 - cos(α)):

\displaystyle \frac{\cos\alpha(1-\cos\alpha)}{(1+\cos\alpha)(1-\cos\alpha)}-\frac{\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha }{\sin^2\alpha }-\frac{1}{\sec\alpha -1}

Simplify:

\displaystyle \frac{\cos\alpha(1-\cos\alpha)}{1-\cos^2\alpha}-\frac{\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha }{\sin^2\alpha }-\frac{1}{\sec\alpha -1}

From the Pythagorean Identity, we know that cos²(α) + sin²(α) = 1 or equivalently, 1 - cos²(α) = sin²(α). Substitute:

\displaystyle \frac{\cos\alpha(1-\cos\alpha)}{\sin^2\alpha}-\frac{\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha }{\sin^2\alpha }-\frac{1}{\sec\alpha -1}

Subtract:

\displaystyle \frac{\cos\alpha(1-\cos\alpha)-\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}

Distribute:

\displaystyle \frac{\cos\alpha-\cos^2\alpha-\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}

Rewrite:

\displaystyle \frac{(\cos\alpha)-(\cos^2\alpha+\cos\alpha)}{\sin^2\alpha}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}

Split:

\displaystyle \frac{\cos\alpha}{\sin^2\alpha}-\frac{\cos^2\alpha+\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}

Factor the second fraction, and substitute sin²(α) for 1 - cos²(α):

\displaystyle \frac{\cos\alpha}{\sin^2\alpha}-\frac{\cos\alpha(\cos\alpha+1)}{1-\cos^2\alpha}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}

Factor:

\displaystyle \frac{\cos\alpha}{\sin^2\alpha}-\frac{\cos\alpha(\cos\alpha+1)}{(1-\cos\alpha)(1+\cos\alpha)}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}

Cancel:

\displaystyle \frac{\cos\alpha}{\sin^2\alpha}-\frac{\cos\alpha}{(1-\cos\alpha)}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}

Divide the second fraction by cos(α):

\displaystyle \frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}=\displaystyle \frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}

Hence proven.

7 0
3 years ago
Write the word sentence as an inequality<br> A number y divided by 6 is no more than 2
Sedbober [7]

Answer:

Y/6 <_ 2

Step-by-step explanation:

(Less than it equal to sign)

8 0
3 years ago
Classify the following triangle as acute, obtuse, or right.
boyakko [2]

Answer:

Acute Triangle.

Explanation:

The figure shows an equilateral triangle which has all the angles equal.

Acute Triangles are those who has angles less than 90°

Hence, an equilateral triangle is always an acute triangle.

Additional,

A right angle triangle has 90° angle.

An obtuse triangle has greater than 90° angle.

6 0
2 years ago
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