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3 years ago

What is y=-3/10x-8 written in standard form?

Mathematics

2 answers:

damaskus [11]3 years ago

7 0

Step-by-step explanation:

Standard form is Ax+By=C

Get the x term to the left side (Add 3/10 x to both sides): y+3/10 x=-8

Multiply by the least common denominator (multiply both sides by 10): 10y+3x=-80

Rewrite in the proper order: 3x+10y=-80

ahrayia [7]3 years ago

5 0

You need to add 3/4x to the side of the y and then it becomes positive and then the lyres side you have only the -8 so it will be 3/4x+y=-8

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HELPPPPP!!! ASAPPPPP!!!!!

ololo11 [35]

Answer:

12 $\sqrt{3}$ / 24

Step-by-step explanation:

ΔXZW is a 30-60-90 triangle whose sides are in the ratio of 1 : $\sqrt{3}$ : 2, respectively

also, ΔXZW ≅ ΔYZW

so XW = 12

XZ = 24

WZ = 12 $\sqrt{3}$

cos ratio of ∡XZW is WZ/XZ which is:

12 $\sqrt{3}$ / 24

3 0

3 years ago

Find two integers whose sum is 0 and product is -64

romanna [79]

Answer:

-8 and 8

Step-by-step explanation:

-8*8 = -64

-8+8=0

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3 years ago

Henry steps onto an elevator at the first floor. He goes up to the 9th floor and then comes down 2 floors. He then goes up 4 flo

dimaraw [331]

Answer:

Step-by-step explanation:

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3 years ago

I need help with number 6.

wolverine [178]

Answer:

(ab+10+a)-(6ab×6ab-2ab+8)

=ab+10+a-6ab×6ab+ab-8

=ab+10+a-36a²b²+2ab-8

=36a²b²+(ab-2ab)+a+(10-8)

therefore the answer is 36a²b²+3b+a+2

4 0

3 years ago

Read 2 more answers

10 marbles what is the probability of drawing 2 blue marbles if the first marble is placed back in the bag before the second dra

stealth61 [152]

Complete Question:

There are 10 marbles. What is the probability of drawing 2 blue marbles if the first marble is NOT placed back in the bag before the second draw.

2 yellow, 3 pink, 5 blue

Answer:

$P(B_1\ and\ B_2) = \frac{2}{9}$

Step-by-step explanation:

Given

$Blue = 5$

$Pink = 3$

$Yellow = 2$

$Total = 10$

When the first blue marble is selected, the probability is:

$P(B_1) = \frac{n(Blue)}{Total}$

$P(B_1) = \frac{5}{10}$

Since the first marble is not returned, we have the following marbles left

$Blue = 4$

$Pink = 3$

$Yellow = 2$

$Total = 9$

The probability of picking a second blue marble is:

$P(B_2) = \frac{n(Blue)}{Total}$

$P(B_1) = \frac{4}{9}$

So, the required probability is:

$P(B_1\ and\ B_2) = P(B_1) * P(B_2)$

$P(B_1\ and\ B_2) = \frac{5}{10} * \frac{4}{9}$

$P(B_1\ and\ B_2) = \frac{5*4}{10*9}$

$P(B_1\ and\ B_2) = \frac{20}{90}$

$P(B_1\ and\ B_2) = \frac{2}{9}$

5 0

3 years ago