All categories

3 years ago

-7-7 (2m-7) = 7 (6+m)

Mathematics

2 answers:

Yakvenalex [24]3 years ago

7 0

Answer:

or,-14(2m-7)=7(6+m)

or,-28m + 98=42 + 7m

or,98-42=7m+28

or,56=35m

or,56/35=m

therefore,m=1.6 Ans

levacccp [35]3 years ago

4 0

Answer:

m= 0

Step-by-step explanation:

this is the right answer

-7-14m+49=42+7m

42-14m=42+7m

14m+7m=42-42

21m=0

m=0

You might be interested in

Can someone please tell me that answer in simplest form

Bezzdna [24]

The answer in simplest for would be 7/8. What you basically do is multiply 7/10 by the reciprocal of 4/5 and that gives u the answer.

5 0

4 years ago

Read 2 more answers

How to solve radical numerator fractions

Nonamiya [84]

Y=radical 3
just cancel out the 2 on both sides

3 0

3 years ago

What is the geometric mean of 4 and 10

GalinKa [24]

Answer: 6.3

Step-by-step explanation:

The formula for geometric mean is given as :

G.m = $\sqrt{pq}$

Therefore , the geometric mean between 4 and 10 will be :

G.m = $\sqrt{4(10)}$

= $\sqrt{40}$

= 2 $\sqrt{10}$

= 6.3

8 0

4 years ago

Whats A and B <br> How do i solve this ?

Aleonysh [2.5K]

Answer:

$\begin{pmatrix}-2\\-4\end{pmatrix}$

Step-by-step explanation:

The question is, how many x and y steps do I have to move the P picture to get it in the Q position. The answer is 2 to the left, so -2, and 4 down, so -4.

4 0

3 years ago

A box with a square base and an open top is being constructed out of A cm2 of material. If the volume of the box is to be maximi

viktelen [127]

Answer:

Side length = $\sqrt{\frac{A}{3} }$ cm , Height = $\frac{1}{2} \sqrt{\frac{A}{3} }$ cm , Volume = $\frac{A\sqrt{A}}{6\sqrt{3} }$ cm³

Step-by-step explanation:

Assume

Side length of base = x

Height of box = y

total material required to construct box = A ( given in question)

So it can be written as

A = x² + 4xy

4xy = A - x²

$y = \frac{A - x^{2} }{4x}$

Volume of box = Area x height

V = x² ₓ y

V = x² ₓ ( $\frac{A - x^{2} }{4x}$ )

V = $\frac{Ax - x^{3} }{4}$

To find max volume put V' = 0

So taking derivative equation becomes

$\frac{A - 3 x^{2} }{4} = 0$

A = 3 $x^{2}$

$x^{2}$ = $\frac{A}{3}$

x = $\sqrt{\frac{A}{3\\} }$

put value of x in equation 1

y = $\frac{A - \frac{A}{3} }{4\sqrt{\frac{A}{3} } }$

y = $\frac{2 \sqrt{\frac{A}{3} } }{4 \sqrt{\frac{A}{3} } }$

y = $\frac{1}{2} \sqrt{\frac{A}{3} }$

So the volume will be

V = $x^{2}$ × y

Put values of x and y from equation 2 & 3

V = $\frac{A}{3} (\frac{1}{2} \sqrt{\frac{A}{3} } )$

V = $\frac{A\sqrt{A}}{6\sqrt{3} }$

8 0

4 years ago