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2 years ago

Multiply. (5v+6w-7)(-3) Simplify your answer.

Mathematics

1 answer:

Zielflug [23.3K]2 years ago

3 0

Answer:

if my calculator is not wrong Is 21

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Need help with this please

Crazy boy [7]

Answer:

1. 2(k + 3)

2. 3. 75 + k

Step-by-step explanation:

1. 2x + 6

Since, 2 is a common factor of 2 and 6, we can take that common outside.

So, 2x + 6 = 2(x + 3)

Note that in the initial expression, this 2 was distributed to arrive at 2x + 6.

2. (1.5 + k) + 2.25

This is simple addition. We can simply remove the brackets to have:

1.5 + k + 2.25

Since, the like terms can be added, we will have:

1.5 + 2. 25 + k

= 3. 75 + k

8 0

3 years ago

14=h/5 Help me please

ValentinkaMS [17]

Answer:

14=h/5

h=14*5

h=70

Check:

70/5=14

Hope this helps

6 0

3 years ago

Evaluate 3a+15+bc−6, when a=7, b=3, and c=15. Enter your answer in the box.

melamori03 [73]

I think it's 75.
____________

5 0

3 years ago

Read 2 more answers

Hong drove 648 miles in 9 hours.

ioda

Answer:

864 miles

Step-by-step explanation:

648/9 = 72 mph

72 x 12 = 864

6 0

3 years ago

Read 2 more answers

Rewrite the expression in terms of the given function 1/1-sinx - sinx/1+sinx

Feliz [49]

Your question seems a bit incomplete, but for starters you can write

$\dfrac1{1-\sin x}-\dfrac{\sin x}{1+\sin x}=\dfrac{1+\sin x}{(1-\sin x)(1+\sin x)}-\dfrac{\sin x(1-\sin x)}{(1+\sin x)(1-\sin x)}=\dfrac{1+\sin x-\sin x(1-\sin x)}{(1-\sin x)(1+\sin x)}$

Expanding where necessary, recalling that $(1-\sin x)(1+\sin x)=1-\sin^2x=\cos^2x$ , you have

$\dfrac{1+\sin x-\sin x(1-\sin x)}{(1-\sin x)(1+\sin x)}=\dfrac{1+\sin x-\sin x+\sin^2x}{\cos^2x}=\dfrac{1+\sin^2x}{\cos^2x}$

and you can stop there, or continue to rewrite in terms of the reciprocal functions,

$\dfrac{1+\sin^2x}{\cos^2x}=\sec^2x+\tan^2x$

Now, since $1+\tan^2x=\sec^2x$ , the final form could also take

$\sec^2x+\tan^2x=\sec^2x+(\sec^2x-1)=2\sec^2x-1$

or

$\sec^2x+\tan^2x=(1+\tan^2x)+\tan^2x=1+2\tan^2x$

7 0

3 years ago