All categories

4 years ago

Solve for x 402)2x - 3 = 60 Round to 3 places

Mathematics

1 answer:

TEA [102]4 years ago

7 0

2x - 3 = 60

(Add 3 to both sides)

2x - 3 + 3 = 60 + 3

2x = 63

(Divide by 2)

2x / 2 = 63 / 2

x = 31.5

You might be interested in

I need so much help please somebody help me

marysya [2.9K]

Answer:

$448

Step-by-step explanation:

You need to find 2% of 400.00, so you would do this: 0.02 · 400 = 8

That means $8 is 2% of 400, so all you need to do is multiply 8 by 6 to get 48. Add that number to the original $400, and you get $448.

<em>Hope this answers your question!</em>

6 0

3 years ago

T-Shirts & More Print Shop will print any image on a coffee mug for a cost of $2 per mug and a one-time charge of $12 to set

Sveta_85 [38]

Answer:

2x+12=5x

Step-by-step explanation:

5 0

4 years ago

Read 2 more answers

Solve the system using either Gaussian elimination with back-substitution or Gauss-Jordan elimination. (If there is no solution,

Gnom [1K]

Answer:

x1=-4/5, x2=18/5 and x3=7/5

Step-by-step explanation:

$\left[\begin{array}{ccc|c}1&0&-3&-5\\3&1&2&4\\2&2&1&7\end{array}\right]$

you can do linear combination between the rows:

2nd row=R2-3R1 and 3th row=R3-2R1

$\left[\begin{array}{ccc|c}1&0&-3&-5\\0&1&11&19\\0&2&7&17\end{array}\right]$

3th row=(3R2-R3)/15

$\left[\begin{array}{ccc|c}1&0&-3&-5\\0&1&11&19\\0&0&1&\frac{7}{5} \end{array}\right]$

1st row=R1+3R3 and R2-11R3

$\left[\begin{array}{ccc|c}1&0&0&-4/5 \\0&1&0&18/5 \\0&0&1&7/5\end{array}\right]$

x1=-4/5, x2=18/5 and x3=7/5

6 0

3 years ago

I need help with this question fast! i don't have a lot of time! can someone help me please???

jarptica [38.1K]

16x^4-81y^4 difference of perfect squares...

(4x^2-9y^2)(4x^2+9y^2) that's one of the equivalents...

(2x+3y)(2x-3y)(4x^2+9y^2) that's another one of them...

5 0

3 years ago

TRIGONOMETRY , can you solve this plz:(<br>even if you know one of them,plz help:)

daser333 [38]

Answer:

See below

Step-by-step explanation:

$\frac{1}{1-cos A} - \frac{1}{1+cos A} =2 cosec A cot A \\ \\ L. H. S.= \frac{1}{1-cos A} - \frac{1}{1+cos A} \\ \\ = \frac{1+cos A - (1 - cos A)}{(1-cos A)(1+cos A)}\\ \\ = \frac{1+cos A - 1 + cos A}{1-cos^{2} A}\\ \\ = \frac{2cos A }{sin^{2} A}\\ \\ =2 .\frac{1}{sin A}.\frac{cos A }{sin A}.\\ \\ = 2.cosec \: A.cot \: A \\ \\ = R.H.S$

5 0

3 years ago