All categories

3 years ago

Solve for d. 41=12d-7 d= ?

Mathematics

2 answers:

gladu [14]3 years ago

4 0

41=12d-7

-12d= -7-41

-12d= -48

d = 4

joja [24]3 years ago

4 0

Answer: d = 4

Step-by-step explanation:

First, switch sides :

12d -7= 41

Then, Add 7 to both sides:

12d- 7 + 7 = 41 +7

After, Simplify :

12d = 48

Next, Divide both sides by 12 :

$\frac{12d}{12} = \frac{48}{12}$

Finally, you get your answer: 

d = 4

* Hopefully this helps:) mark me the brainliest:) !!

∞ 234483279c20∞

You might be interested in

Match the expressions with their equivalent simplified expressions.

Tasya [4]

Answer:

$\sqrt[4]{\frac{16x^6y^4}{81x^2y^8}}\rightarrow\frac{2x}{3y}\\\sqrt[4]{\frac{81x^2y^{10}}{81x^6y^6}} \rightarrow\frac{3y}{2x}\\\sqrt[3]{\frac{64x^8y^7}{125x^2y^{10}}}\rightarrow\frac{4x^2}{5y}\\\sqrt[5]{\frac{243x^{17}y^{16}}{32x^7y^{21}}}\rightarrow\frac{3x^2}{2y}\\\sqrt[5]{\frac{32x^{12}y^{15}}{243x^7y^{10}}} \rightarrow\frac{2xy}{3}\\\sqrt[4]{\frac{16x^{10}y^{9}}{256x^2y^{17}}}\rightarrow\frac{x}{2y}$

Step-by-step explanation:

$\sqrt[4]{\frac{16x^6y^4}{81x^2y^8}} =\sqrt[4]{\frac{(2^4)(x^{6-2})(y^{4-8})}{(3^4)}} =\sqrt[4]{\frac{2^4x^4y^{-4}}{3^4}} =\frac{2xy^{-1}}{3}=\frac{2x}{3y}$

$\sqrt[4]{\frac{81x^2y^{10}}{81x^6y^6}} =\sqrt[4]{\frac{(3^4)(x^{2-6})(y^{10-6})}{(2^4)}} =\sqrt[4]{\frac{3^4x^{-4}y^{4}}{2^4}} =\frac{3x^{-1}y^1}{3}=\frac{3y}{2x}$

$\sqrt[3]{\frac{64x^8y^7}{125x^2y^{10}}} =\sqrt[3]{\frac{(4^3)(x^{8-2})(y^{7-10})}{(5^3)}} =\sqrt[3]{\frac{4^3x^6y^{-3}}{5^3}} =\frac{4x^2y^{-1}}{5}=\frac{4x^2}{5y}$

$\sqrt[5]{\frac{243x^{17}y^{16}}{32x^7y^{21}}} =\sqrt[5]{\frac{(3^5)(x^{17-7})(y^{16-21})}{(2^5)}} =\sqrt[5]{\frac{3^5x^{10}y^{-5}}{2^5}} =\frac{3x^2y^{-1}}{2}=\frac{3x^2}{2y}$

$\sqrt[5]{\frac{32x^{12}y^{15}}{243x^7y^{10}}} =\sqrt[5]{\frac{(2^5)(x^{12-7})(y^{15-10})}{(3^5)}} =\sqrt[5]{\frac{2^5x^{5}y^{5}}{3^5}} =\frac{2x^1y^{1}}{3}=\frac{2xy}{3}$

$\sqrt[4]{\frac{16x^{10}y^{9}}{256x^2y^{17}}} =\sqrt[4]{\frac{(2^4)(x^{10-2})(y^{9-17})}{(4^4)}} =\sqrt[4]{\frac{2^4x^{8}y^{-8}}{4^4}} =\frac{2x^{1}y^{-1}}{4}=\frac{x}{2y}$

Thus,

3 0

3 years ago

If 2k, 5k-1 and 6k+2 are the first 3 terms of an arithmetic sequence, find k and the 8th term.

Simora [160]

Consecutive terms in an arithmetic sequence differ by a constant $d$ . So

$5k-1=2k+d$
$6k+2=5k-1+d$
$\implies\begin{cases}3k-d=1\\k-d=-3\end{cases}\implies k=2,d=5$

Denote the $n$ -th term in the sequence by $a_n$ . Now that $a_1=2$ , we have

$a_n=a_{n-1}+5=a_{n-2}+2\cdot5=\cdots=a_1+(n-1)\cdot5$

which means

$a_8=2+(8-1)\cdot5=37$

4 0

2 years ago

Which decimal has the greatest value?

cupoosta [38]

Answer:

Answer: D

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

What is the relationship between 1 and 2? WILL GIVE BRAINLYEST.

netineya [11]

Answer:

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

I really dont understand what to do here if you can please help

NNADVOKAT [17]

Answer:

see below

Step-by-step explanation:

The two angles form a straight line so they add to 180 degrees

5x+60 + 3x-40 = 180

Combine like terms

8x +20 = 180

Subtract 20 from each side

8x+20-20=180-20

8x= 160

Divide by 8

8x/8=160/8

x=20

5x+60 = 5*20+60 = 160

3x-40 = 3*20-40 = 20

5 0

2 years ago