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3 years ago

Write the coordinates of the point 2 units up of the point (-5,7)

Mathematics

1 answer:

babymother [125]3 years ago

3 0

Answer:

(-5,9)

Step-by-step explanation:

7 + 2 is 9 so yeah that's it

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2x - x + 3x + 5 = 25 Solve for X

postnew [5]

Answer:

$\bf x=5$

Step-by-step explanation:

$\bf 2x - x + 3x + 5 = 25$

Combine like terms.

$\bf 4x+5=25$

Subtract 5 from both sides.

$\bf 4x+5-5=25-5$

$\bf 4x=20$

Divide both sides by 4.

$\bf \cfrac{4x}{4}=\cfrac{20}{4}$

$\bf x=5$

7 0

2 years ago

Read 2 more answers

Need help asap!!

In-s [12.5K]

Answer:D

Step-by-step explanation:

(3x^3 - 5x^2 + 4x - 9)-(7x^3 - 8x^2 - 5x + 10)

open brackets

3x^3 - 5x^2 + 4x - 9 - 7x^3 + 8x^2 +5x - 10

Collect like terms

3x^3-7x^3-5x^2+8x^2+4x+5x-9-10

-4x^3+3x^2+9x-19

8 0

3 years ago

Which of the following expressions “ the sum of n and the sum of n and 6 ?

faltersainse [42]

what do u mean by that

3 0

3 years ago

If 3v/7=6, then v= What is the Answer to the problem?

Dmitrij [34]

Let's solve this problem step-by-step:

$\frac{3v}{7} =6\\\frac{3v}{7} *1=6\\\frac{3v}{7} *\frac{7}{7}=6\\ 3v = 6 * 7\\3v=42\\v=14$

Thus v = 14

Answer: v= 14

Hope that helps!

5 0

2 years ago

Finding Derivatives Implicity In Exercise,Find dy/dx implicity. x2e - x + 2y2 - xy = 0

Klio2033 [76]

Answer:

the question is incomplete, the complete question is

"Finding Derivatives Implicity In Exercise,Find dy/dx implicity . $x^{2}e^{-x}+2y^{2}-xy$ "

Answer : $\frac{dy}{dx}=\frac{y-(2-x)xe^{-x}}{(4y-x)}$

Step-by-step explanation:

From the expression $x^{2}e^{-x}+2y^{2}-xy$ " y is define as an implicit function of x, hence we differentiate each term of the equation with respect to x.

we arrive at

$\frac{d}{dx}(x^{2}e^{-x )+\frac{d}{dx} (2y^{2})-\frac{d}{dx}xy=0\\$

for the expression $\frac{d}{dx}(x^{2}e^{-x})$ we differentiate using the product rule, also since y^2 is a function of y which itself is a function of x, we have

$(2xe^{-x}-x^{2}e^{-x})+4y\frac{dy}{dx}-x\frac{dy}{dx} -y=0\\\\(2-x)xe^{-x}+(4y-x)\frac{dy}{dx}-y=0 \\$ .

if we make dy/dx subject of formula we arrive at

$(4y-x)\frac{dy}{dx}=y-(2-x)xe^{-x}\\\frac{dy}{dx}=\frac{y-(2-x)xe^{-x}}{(4y-x)}$

5 0

4 years ago