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1 year ago

Find the slope on the line

Mathematics

1 answer:

Alla [95]1 year ago

8 0

Answer:

-1

Step-by-step explanation:

Use Rise/Run and since the line is negative the slope will be negative

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A school replaced 20% of it computers with new ones . what the total number of computers in the school if 55 computers were repl

sladkih [1.3K]

The total is 275 because 100% divided by 20% is 5 and if 20%= 55 then that means that 55 goes into 100% 5 times so you multiply 55 by 5 to get 275. Hope I helped! Please rate me as the brainliest!

7 0

3 years ago

I need help finding the answers someone plz help

Harlamova29_29 [7]

Happy New Year from MrBillDoesMath!

Answer:

Proof by ASA congruence postulate. See below

Discussion:

Fact 1 : angle A = angle T (given)

Fact 2: The angles on both sides of point X are equal as vertical angles

are equal.

From these facts it follows that angle M = angle H (as all plane triangles have 180 degrees). Also AM = TH (given) so

In the left triangle In the right triangle

(angle M, side AM, angle A) = (angle H, side TH, angle T)

Hence the triangles have two congruent angles, and congruent sides included between the angles, so they are congruent by ASA.

Thank you,

MrB

8 0

2 years ago

Need help ASAP to this math question about how many gallons it will take to fill a pool

NARA [144]

1 cubic foot = 7.48 gallons

hence
$\frac{7.48 gal}{1 ft^3} =1$

unit conversion
$\frac{3000 ft^3}{} \times \frac{7.48 gal}{1 ft^3} = 3000 \times 7.48 gal$

3000 x 7.48 =22440

3 0

2 years ago

Prove that: (b²-c²/a)CosA+(c²-a²/b)CosB+(a²-b²/c)CosC = 0

IRISSAK [1]

Prove that:

$\:\:\sf\:\:\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C=0$

Proof: 

We know that, by Law of Cosines,

$\sf \cos A=\dfrac{b^2+c^2-a^2}{2bc}$
$\sf \cos B=\dfrac{c^2+a^2-b^2}{2ca}$
$\sf \cos C=\dfrac{a^2+b^2-c^2}{2ab}$

Taking LHS

$\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C$

Substituting the value of cos A, cos B and cos C,

$\longmapsto\left(\dfrac{b^2-c^2}{a}\right)\left(\dfrac{b^2+c^2-a^2}{2bc}\right)+\left(\dfrac{c^2-a^2}{b}\right)\left(\dfrac{c^2+a^2-b^2}{2ca}\right)+\left(\dfrac{a^2-b^2}{c}\right)\left(\dfrac{a^2+b^2-c^2}{2ab}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2-a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2-b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2-c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2)-(b^2-c^2)(a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2)-(c^2-a^2)(b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2)-(a^2-b^2)(c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^4-c^4)-(a^2b^2-a^2c^2)}{2abc}\right)+\left(\dfrac{(c^4-a^4)-(b^2c^2-a^2b^2)}{2abc}\right)+\left(\dfrac{(a^4-b^4)-(a^2c^2-b^2c^2)}{2abc}\right)$