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3 years ago

−(−10−6)= And explain how you did it 20 points

Mathematics

1 answer:

kolezko [41]3 years ago

3 0

-(-10-6)=x
if there is nothing behind the negative sign put a 1 there.
-1(-10-6)=x
-1*(-10)= 10
-1*6= -6
10-(-6)= 16

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Combine the like terms to create an equivalent expression: <br> 2s+(−4s)=?

Sergeeva-Olga [200]

Answer:

$2s+\left(-4s\right)=-2s$ is the equivalent expression as $2s+\left(-4s\right)$ is equal to $-2s$ .

Step-by-step explanation:

Considering the expression

$2s+\left(-4s\right)$

As we know that

Like terms are said to be the terms which hold the same variables raised to the same power.

For example, $3x$ and $2x$ are like terms as they contain the same variable $x$ raised to the same power.

Lets solve the expression

$2s+\left(-4s\right)$

$\mathrm{Remove\:parentheses}:\quad \left(-a\right)=-a$

$2s-4s$

$\mathrm{Add\:similar\:terms:}\:2s-4s=-2s$

$-2s$

Therefore,

$2s+\left(-4s\right)=-2s$ is the equivalent expression as $2s+\left(-4s\right)$ is equal to $-2s$ .

Keywords: equivalent expression, like terms

Learn more about equivalent expression from brainly.com/question/8902137

#learnwithBrainly

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3 years ago

In ΔABC (m∠C = 90°), the points D and E are the points where the angle bisectors of ∠A and ∠B intersect respectively sides BC an

Airida [17]

This is a little long, but it gets you there.

ΔEBH ≅ ΔEBC . . . . HA theorem
EH ≅ EC . . . . . . . . . CPCTC
∠ECH ≅ ∠EHC . . . base angles of isosceles ΔEHC
ΔAHE ~ ΔDGB ~ ΔACB . . . . AA similarity
∠AEH ≅ ∠ABC . . . corresponding angles of similar triangle
∠AEH = ∠ECH + ∠EHC = 2∠ECH . . . external angle is equal to the sum of opposite internal angles (of ΔECH)
ΔDAC ≅ ΔDAG . . . HA theorem
DC ≅ DG . . . . . . . . . CPCTC
∠DCG ≅ ∠DGC . . . base angles of isosceles ΔDGC
∠BDG ≅ ∠BAC . . . .corresponding angles of similar triangles
∠BDG = ∠DCG + ∠DGC = 2∠DCG . . . external angle is equal to the sum of opposite internal angles (of ΔDCG)
∠BAC + ∠ACB + ∠ABC = 180° . . . . sum of angles of a triangle
(∠BAC)/2 + (∠ACB)/2 + (∠ABC)/2 = 90° . . . . division property of equality (divide equation of 12 by 2)
∠DCG + 45° + ∠ECH = 90° . . . . substitute (∠BAC)/2 = (∠BDG)/2 = ∠DCG (from 10 and 11); substitute (∠ABC)/2 = (∠AEH)/2 = ∠ECH (from 5 and 6)
This equation represents the sum of angles at point C: ∠DCG + ∠HCG + ∠ECH = 90°, ∴ ∠HCG = 45° . . . . subtraction property of equality, transitive property of equality. (Subtract ∠DCG+∠ECH from both equations (14 and 15).)

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soldier1979 [14.2K]

Answer:

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Step-by-step explanation:

80.50 plus........plus what?

i am confusion

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