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stira [4]
2 years ago
7

Can anyone help me with this. It asks to solve and check your answer.

Mathematics
1 answer:
tankabanditka [31]2 years ago
6 0
\sf 4m-21=3(5m-1)

First, distribute 3 into the parenthesis(multiply it to every term inside):

\sf 4m-21=15m-3

Add 21 to both sides:

\sf 4m=15m+18

Subtract 15m to both sides:

\sf -11m=18

Divide -11 to both sides:

\sf m=-\dfrac{18}{11}

To check our work we can plug this back into the original equation for 'm':

\sf 4m-21=3(5m-1)

\sf 4(-\dfrac{18}{11})-21=3(5(-\dfrac{18}{11})-1)

Multiply:

\sf -\dfrac{72}{11}-21=3(-\dfrac{90}{11}-1)

Distribute 3 into the parenthesis:

\sf -\dfrac{72}{11}-21=-\dfrac{270}{11}-3

Subtract:

\sf -\dfrac{303}{11}=-\dfrac{303}{11}~\checkmark

Both sides are equal to each other, so we solved it correctly.
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A:7.031

Step-by-step explanation:

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The overall distance traveled by a golf ball is tested by hitting the ball with Iron Byron, a mechanical golfer with a swing tha
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GEOMETRY PROOFS!
yarga [219]

From the given figure ,

RECA is a quadrilateral

RC divides it into two parts

From the triangles , ∆REC and ∆RAC

RE = RA (Given)

angle CRE = angle CRA (Given)

RC = RC (Common side)

Therefore, ∆REC is Congruent to ∆RAC

∆REC =~ ∆RAC by SAS Property

⇛CE = CA (Congruent parts in a congruent triangles)

Hence , Proved

<em>Additional</em><em> comment</em><em>:</em><em>-</em>

SAS property:-

"The two sides and included angle of one triangle are equal to the two sides and included angle then the two triangles are Congruent and this property is called SAS Property (Side -Angle-Side)

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4 0
2 years ago
Read 2 more answers
The base of an isosceles triangle is one and a half times the length of the other two sides. A smaller triangle has a perimeter
stira [4]

Answer:

P = 1.5x.

Step-by-step explanation:

Let the length of each side of the first triangle be x, then

Perimeter of this triangle

= 2x + 1.5x

= 3.5x.

So perimeter of smaller triangle is:

P = 1/2 * 3.5x

P = 1,.5x.

5 0
3 years ago
6. a semicircle has as its diameter the hypotenuse of a right triangle shown below. determine the area of the semicircle to the
jeyben [28]

Answer:

A = 137.3cm^2

Step-by-step explanation:

Given

See attachment

Required

The area of the semicircle

First, we calculate the hypotenuse (h) of the triangle

Considering only the triangle, we have:

\cos(68) = \frac{7}{h} --- cosine formula

Make h the subject

h = \frac{7}{\cos(68)}

h = \frac{7}{0.3746}

h = 18.7

The area of the semicircle is then calculated as:

A = \frac{\pi h^2}{8}

This gives:

A = \frac{3.14 * 18.7^2}{8}

A = \frac{1098.03}{8}

A = 137.3cm^2

7 0
3 years ago
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