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3 years ago

Quick anyone know the answer to this problem? Will give brainiest

Mathematics

2 answers:

natka813 [3]3 years ago

5 0

Answer:

I think its b.

Natasha2012 [34]3 years ago

4 0

Answer:

c, I think

Step-by-step explanation:

I think its see because you already eliminate b and d because they make no sense and that leaves you with a and c but the problem is a fraction so a is eliminated so I think it would be c, check just in case

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The temperature outside changed from 79°F to 44°F over a period of five days. If the temperature changeu

bixtya [17]

Answer:

B: -7°F

Step-by-step explanation:

79-44=35

Because its over a period of five days we would divide

35÷5=7

Since the numbers are going down its negative

-7

5 0

3 years ago

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Can someone help me with this?

vovikov84 [41]

Answer:

1. obtuse, has an angle with more than 90

2. Acute, all angles are less than 90

3. right has and angle with 90

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3 years ago

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Please help me I need help please

MakcuM [25]

Answer:

p = 4

Step-by-step explanation:

Given equation:

$x^2+(p-3)y^2-4x+6y-16=0$

<u>Standard equation of a circle:</u>

$(x-a)^2+(y-b)^2=r^2$

(where $(a,b)$ is the centre of the circle, and $r$ is the radius)

If you expand this equation, you will see that the coefficient of $y^2$ is always one.

Therefore, $p-3=1$

$\implies p=1+3=4$

<u>Additional information</u>

To rewrite the given equation in the standard form:

$\implies x^2+y^2-4x+6y-16=0$

$\implies x^2-4x+y^2+6y=16$

$\implies (x-2)^2-4+(y+3)^2-9=16$

$\implies (x-2)^2+(y+3)^2=16+4+9$

$\implies (x-2)^2+(y+3)^2=29$

So this is a circle with centre (2, -3) and radius √29

3 0

2 years ago

Study the solutions of three equations on the right. Then complete the statements below.

Hoochie [10]

Answer:

Part 1) There are two real solutions

Part 2) There is one real solution

Part 3) There are no real solutions

Step-by-step explanation:

we know that

The formula to solve a quadratic equation of the form $ax^{2} +bx+c=0$ is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

If the radicand is positive

$(b^{2}-4ac) >0$ ------> There are two real solutions

If the radicand is zero

$(b^{2}-4ac) =0$ ------> There is one real solution

If the radicand is negative

$(b^{2}-4ac)$ ------> There are no real solutions

therefore

Part 1) The radicand is positive

There are two real solutions

Part 2) The radicand is equal to zero

There is one real solution

Part 3) The radicand is negative

There are no real solutions

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3 years ago

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What does it mean by 2cm:1unit on a graph

Alexxx [7]

That means that for every 2cm you measure in real life, you represent it as 1 unit on the graph

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3 years ago