All categories

2 years ago

6y-5=11 what the answer I’m having a hard time finding it

Mathematics

2 answers:

QveST [7]2 years ago

6 0

6y-5=11

Move -5 to the other side. Sign changes from -5 to +5.

6y-5+5=11+5

6y=11+5

6y=16

Divide by 6 for both sides to get y by itself.

6y/6=16/6

Cross out 6 and 6, divide by 6 and then becomes 1*1*y=y

y=16/6

Reduce 16/6 by dividing by 2

16/2=8

6/2=3

Answer: y=8/3 or y=2 2/3

monitta2 years ago

6 0

Add 5 on each side leaving you with 6y=16. Then divide by 6 on each side leaving you with y=16/6. Reducing it gives you y=8/3

You might be interested in

I can’t remember how to do this if I try a bit longer I can probably figure it out but not sure exactly. Please If anyone can he

UNO [17]

Answer:

£2.38

Step-by-step explanation:

2.5 cost £1.40

4.25 cost £ x

so x=(4.25*1.40)/2.5=5.95/2.5=£2.38

5 0

2 years ago

Probability question. Please answer with work attached.

musickatia [10]

Answer:

B. (0.61)

Step-by-step explanation:

Well, first I created venn diagram to help separate my work, then I subtracted .22 from each "probability sections" then I added them up to get .61

Work explains a lot.

Hope it helped. :)

8 0

3 years ago

Is -6 over 7 a whole number a integer number or a rational number

vodka [1.7K]

Answer:

It is a rational number

Step-by-step explanation:

It is a rational number because -6/7 is equal to -0.857142 repeating. And a rational number repeats.

Hope this helps☝️☝☝

6 0

3 years ago

Read 2 more answers

Please help me answer this question mathematicians

Damm [24]

Answer:

6.5e^(i·(-157.38°)) ≈ 6.5e^(-2.7468i)

Step-by-step explanation:

A suitable calculator can find the value of this ratio for you.

((2+3i)/(1-i))² = -6 -2.5i ≈ 6.5∠-157.38° ≈ 6.5e^(-2.7468i)

The second attachments shows the calculator set to radian mode for the angle measure.

6 0

1 year ago

Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab

Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

$SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx$

If $f(x)=x^2,$ then

$f'(x)=2x$

and

$b=0\\ \\a=2$

Therefore,

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx$

Apply substitution

$x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du$

Then

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du$

Now

$\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))$

Hence,

$SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}$

3 0

2 years ago