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Hatshy [7]
3 years ago
12

The base of a triangle is 6 meters longer than the height of the triangle. If the area of the triangle is 108 square meters, wha

t are the base and height of the triangle?
Mathematics
1 answer:
Nonamiya [84]3 years ago
4 0

Answer:

36 meters

Step-by-step explanation:

A=1/2b×h

108sq. m=1/2.6m×h

both side canceling meter

h=2×108m/6

h=36m. answer

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What value(s) of x will make each equation below true
Dahasolnce [82]

We need to solve for x. Let's try problem b:

2x-6=3x+1-x-7

Let us first combine line terms. 3x and -x as well as 1 and -7 can be combined. Let's do that:

2x-6=2x-6

Since this is true, your answer would be:

All real numbers

------------------------------------------------------------------

Let's solve for problem c:

3x+1=43

Let's isolate x, so subtract 1 from both sides:

3x=42

Since x can't have a coefficient, divide both sides by 3:

\frac{3x}{3}=\frac{42}{3}

x=14

So, only the value of 14 would make this equation true.

------------------------------------------------------------------

Let's try problem d:

4x-1=4x+7

Let's get our whole numbers on the right side. Add 1 to both sides:

4x=4x+8

Subtract 4x from the right side on both sides:

0=8

Since this is not true, your answer would be:

No solution

7 0
3 years ago
I spent all my points into this so give it your best shot
ioda

Answer: hi

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
Match the equation with its corresponding solution for x. 6= -1/3x+5
KiRa [710]

Answer:

x=-3

Step-by-step explanation:

7 0
3 years ago
Explain to a new student how a reflection across the y-axis changes the coordinates of the original point.
RSB [31]
If you were to have (3,2) and find the reflection of the y axis, it would turn y into a negative, making it (3,-2).
6 0
3 years ago
Plz Helppp meeee assapppp
Gwar [14]

Given:

A right angled triangle.

To find:

The value of x.

Solution:

We have,

Base = x

Hypotenuse = \sqrt{5}

In a right angle triangle,

\cos \theta = \dfrac{Base}{Hypotenuse}

For the given triangle,

\cos 45^\circ= \dfrac{x}{\sqrt{5}}

\dfrac{1}{\sqrt{2}}= \dfrac{x}{\sqrt{5}}

Multiply both sides by \sqrt{5}.

\dfrac{\sqrt{5}}{\sqrt{2}}= x

\dfrac{\sqrt{5}}{\sqrt{2}}\times \dfrac{\sqrt{2}}{\sqrt{2}}= x

\dfrac{\sqrt{10}}{2}= x

Therefore, the required value of x is \dfrac{\sqrt{10}}{2}.

8 0
2 years ago
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