All categories

3 years ago

Write the equation y+5=-10(x-1) into standard form

Mathematics

1 answer:

olga55 [171]3 years ago

8 0

Answer:

10x + y = 5

General Formulas and Concepts:

Pre-Algebra

Equality Properties

Algebra I

Point-Slope Form: y - y₁ = m(x - x₁)

x₁ - x coordinate

y₁ - y coordinate

m - slope

Standard Form: Ax + By = C

Step-by-step explanation:

Step 1: Define

Point-Slope Form: y + 5 = -10(x - 1)

Step 2: Find Standard Form

Distribute -10: y + 5 = -10x + 10
Add 10x on both sides: 10x + y + 5 = 10
Subtract 5 on both sides: 10x + y = 5

And we have our final answer!

You might be interested in

Pls help I need this done asap

Paul [167]

Answer:

I can help you solve this.

but first telll me in the comments what is the formula for this equation and ill tell u from there

5 0

2 years ago

For the given term, find the binomial raised to the power, whose expansion it came from: 15(5)^2 (-1/2 x) ^4

Elina [12.6K]

Answer:

C. $(5-\frac{1}{2})^6$

Step-by-step explanation:

Given

$15(5)^2(-\frac{1}{2})^4$

Required

Determine which binomial expansion it came from

The first step is to add the powers of he expression in brackets;

$Sum = 2 + 4$

$Sum = 6$

Each term of a binomial expansion are always of the form:

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

Where n = the sum above

$n = 6$

Compare $15(5)^2(-\frac{1}{2})^4$ to the above general form of binomial expansion

$(a+b)^n = ......+15(5)^2(-\frac{1}{2})^4+.......$

Substitute 6 for n

$(a+b)^6 = ......+15(5)^2(-\frac{1}{2})^4+.......$

[Next is to solve for a and b]

From the above expression, the power of (5) is 2

Express 2 as 6 - 4

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

By direct comparison of

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

and

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

We have;

$^nC_ra^{n-r}b^r= 15(5)^{6-4}(-\frac{1}{2})^4$

Further comparison gives

$^nC_r = 15$

$a^{n-r} =(5)^{6-4}$

$b^r= (-\frac{1}{2})^4$

[Solving for a]

By direct comparison of $a^{n-r} =(5)^{6-4}$

$a = 5$

$n = 6$

$r = 4$

[Solving for b]

By direct comparison of $b^r= (-\frac{1}{2})^4$

$r = 4$

$b = \frac{-1}{2}$

Substitute values for a, b, n and r in

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

$(5+\frac{-1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

Solve for $^6C_4$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{(6-4)!4!)}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{2!!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5*4!}{2*1*!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5}{2*1}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{30}{2}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^2(\frac{-1}{2})^4+.......$

Check the list of options for the expression on the left hand side

The correct answer is $(5-\frac{1}{2})^6$

3 0

3 years ago

Answer question please.?

marin [14]

2. 8 books total 2 books fiction= 2/8 fiction

4. 6 ferns total 2 have been watered= 4/6 havent been

6. 3/4=6/8 clamshelss

8 0

3 years ago

What is 8 * 3/4 hour

Pachacha [2.7K]

78 * 3/4 hour,
it is 6 hours

5 0

3 years ago

Read 2 more answers

Merrida uses a pattern in the multiplication table below to find ratios that are equivalent to 7:9. If Merrida multiplies the fi

astra-53 [7]

Answer:

Multiply 9 by 6

Step-by-step explanation:

7:9 is the ratio

We multiply the first term by 6.

What we do to one side, we do to the other

7*6=42

9*6 = 54

That way we keep the ratio in the same proportion

42:54

6 0

3 years ago

Read 2 more answers