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2 years ago

What is the solution to the equation 400 - 8m =216

Mathematics

2 answers:

stepan [7]2 years ago

7 0

Answer:

400−8m=216

Step 1: Simplify both sides of the equation.

400−8m=216

400+−8m=216

−8m+400=216

Step 2: Subtract 400 from both sides.

−8m+400−400=216−400

−8m=−184

Step 3: Divide both sides by -8.

−8m

−8

−184

−8

m=23

777dan777 [17]2 years ago

5 0

Answer: m=23

Step-by-step explanation:

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Find the values of X and Y that makes these triangles congruent by the HL theorem

muminat

Answer:

C. x = 3, y = 2

Step-by-step explanation:

If both triangles are congruent by the HL Theorem, then their hypotenuse and a corresponding leg would be equal to each other.

Thus:

x + 3 = 3y (eqn. 1) => equal hypotenuse

Also,

x = y + 1 (eqn. 2) => equal legs

✔️Substitute x = y + 1 into eqn. 1 to find y.

x + 3 = 3y (eqn. 1)

(y + 1) + 3 = 3y

y + 1 + 3 = 3y

y + 4 = 3y

y + 4 - y = 3y - y

4 = 2y

Divide both sides by 2

4/2 = 2y/2

2 = y

y = 2

✔️ Substitute y = 2 into eqn. 2 to find x.

x = y + 1 (eqn. 2)

x = 2 + 1

x = 3

8 0

2 years ago

What is 485,783 rounded to its greatest place?

natima [27]

Answer:

486

Step-by-step explanation:

if you mean 485.782 with a decimal then it would be 486

7 0

2 years ago

Given:

diamong [38]

Answer:

10-5 $\sqrt{2}$

Step-by-step explanation:

As per the attached figure, right angled $\triangle MDL$ has an inscribed circle whose center is $I$ .

We have joined the incenter $I$ to the vertices of the $\triangle MDL$ .

Sides MD and DL are equal because we are given that $\angle M = \angle L = 45 ^\circ.$

Formula for <em>area</em> of a $\triangle = \dfrac{1}{2} \times base \times height$

As per the figure attached, we are given that side <em>a = 10.</em>

Using pythagoras theorem, we can easily calculate that side ML = 10 $\sqrt{2}$

Points P,Q and R are at $90 ^\circ$ on the sides ML, MD and DL respectively so IQ, IR and IP are heights of $\triangle$ MIL, $\triangle$ MID and $\triangle$ DIL.

Also,

$\text {Area of } \triangle MDL = \text {Area of } \triangle MIL +\text {Area of } \triangle MID+ \text {Area of } \triangle DIL$

$\dfrac{1}{2} \times 10 \times 10 = \dfrac{1}{2} \times r \times 10 + \dfrac{1}{2} \times r \times 10 + \dfrac{1}{2} \times r \times 10\sqrt2\\\Rightarrow r = \dfrac {10}{2+\sqrt2} \\\Rightarrow r = \dfrac{5\sqrt2}{\sqrt2+1}\\\text{Multiplying and divinding by }(\sqrt2 +1)\\\Rightarrow r = 10-5\sqrt2$