All categories

2 years ago

5x + 5y = -5 when y = 1 solve for x in 10 points .

Mathematics

1 answer:

san4es73 [151]2 years ago

4 0

Step-by-step explanation:

5x + 5y = -5 , when y = 1

5x + 5.1 = -5

5x + 5 = -5

5x = -5 - 5

5x = -10

x = -10/5

x = -2

You might be interested in

Find the value.<br> 9x+6 when x=-2/3

OLEGan [10]

Answer:

Step-by-step explanation:

Given,

$- \frac{2}{3}$

Now, let's find the value of 9x + 6

$9x + 6$

plug the value of X

$= 9 \times (- \frac{2}{3} ) + 6$

Multiplying a positive and negative equals a negative

$( + ) \times ( - ) = ( - )$

$= - 9 \times \frac{2}{3} + 6$

Reduce the number with Greatest Common Factor 3

$= - 3 \times 2 + 6$

Multiply the numbers

$= - 6 + 6$

The sum of two opposites equals 0

$= 0$

Hope this helps..

best regards!!

7 0

2 years ago

PLZ HURRY ITS TIMEDDDDD

GaryK [48]

A. 1/5k - 2/3j and -2/3j + 1/5k

Step-by-step explanation:

Equivalent expressions are expressions that are the same even though they may appear different.When you plug in a value to represent a variable in these expression, they will give same answer when simplified.

1/5k - 2/3j and -2/3j + 1/5k you notice the operation signs in the terms has been maintained though the position of the terms shifted.

Learn More

Equivalent expressions :brainly.com/question/280220

Keywords: expressions, equivalent

#LearnwithBrainly

8 0

2 years ago

Read 2 more answers

Find the distance between points p(8, 2) and q(3, 8) to the nearest tenth.

grandymaker [24]

Distance between two points, d, is given by
$d= \sqrt{ ( x_{2} - x_{1}) ^{2} + (y_{2} - y_{1}) ^{2} }
$
where: (x1, y1) = (8, 2) and (x2, y2) = (3, 8)
$d= \sqrt{ ( 3 - 8) ^{2} + (8 - 2) ^{2} } \\ d= \sqrt{ (-5)^{2} + 6^{2} } \\ d= \sqrt{25+36} \\ d= \sqrt{61} \\ d=7.8 \ units$

3 0

2 years ago

Read 2 more answers

Find the taylor polynomial t3(x) for the function f centered at the number

inysia [295]

Answer:

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3$

Step-by-step explanation:

We are given that

$f(x)=7tan^{-1}(x)$

a=1

$T_n(x)=\sum_{r=0}^{n}\frac{f^r(a)(x-a)^r}{r!}$

Substitute n=3 and a=1

$t_3(x)=f(1)+f'(1)(x-1)+\frac{f''(1)(x-1)^2}{2!}+\frac{f'''(1)(x-1)^3}{3!}$

$f(x)=7tan^{-1}(x)$

$f(1)=7tan^{-1}(1)=7\times \frac{\pi}{4}=\frac{7\pi}{4}$

Where $tan^{-1}(1)=\frac{\pi}{4}$

$f'(x)=\frac{7}{1+x^2}$

Using the formula

$\frac{d(tan^{-1}(x))}{dx}=\frac{1}{1+x^2}$

$f'(1)=\frac{7}{2}$

$f''(x)=\frac{-14x}{(1+x^2)^2}$

$f''(1)=-\frac{7}{2}$

$f''(x)=-14x(x^2+1)^{-2}$

$f'''(x)=-14((x^2+1)^{-2}-4x^2(x^2+1)^{-3}})$

By using the formula

$(uv)'=u'v+v'u$

$f'''(x)=-14(\frac{x^2+1-4x^2}{(1+x^2)^3}$

$f'''(x)=(-14)\frac{-3x^2+1}{(1+x^2)^3}$

$f'''(1)=-14(\frac{-3(1)+1}{2^3})=\frac{7}{2}$

Substitute the values

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{2\times 3\times 2\times 1}(x-1)^3$

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3$

7 0

2 years ago

Isabella's house has 3 bedrooms. Each bedroom is 12 feet long, 10 feet wide, and 8 feet high. Isabella must paint the walls of a

san4es73 [151]

Answer:

876 square feet need to be painted.

Step-by-step explanation:

There are four walls in each bedroom, since she can't paint floors or ceilings. So we calculate the number of square feet of wall there is in one bedroom

(12*8)+(12*8)+(10*8)+(10*8)-60=160+192-60=292

There are 3 bedrooms so

292*3=876

6 0

2 years ago