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2 years ago

Find the area of the shaded part of the triangle below. Round to the nearest tenth if needed.

Mathematics

2 answers:

konstantin123 [22]2 years ago

7 0

Answer:

212.5

Step-by-step explanation:

For triangles, A = (bh)/2

You can do 15+25 to find the base, which is 40. From here do 17*40/2. You get 340. However, you need to remove the right triangle on the left side.

Again do 17*15/2, which you get 127.5.

From here subtract 127.5 from 340 and get 212.5

vaieri [72.5K]2 years ago

4 0

Answer:

Step-by-step explanation:

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Help pls, will mark brainliest

BaLLatris [955]

Here , we are provided with a table which shows 5 consecutive terms of an arithmetic sequence . But before solving further , let's recall that ;

The n'th term of a Arithmetic Sequence let's say it be ${\sf T_n}$ is given by ;

${\boxed{\bf T_{n}=T_{1}+(n-1)d}}$

Where , <u>d</u> is the common difference

Now , here we are given with ;

${\quad \qquad \sf \blacktriangleright \blacktriangleright \blacktriangleright T_{1}=8 \: and \: T_{5}=-4}$

We have to find the 2nd , 3rd and 4th term respectively ,

Now , by using the above formula , 5th term can be written as ;

${: \implies \quad \sf T_{1}+(5-1)d=T_{5}}$

Putting the values and transposing 1st term to RHS , we have ;

${: \implies \quad \sf 4d = -4-8}$

${: \implies \quad \sf d=-\dfrac{12}{4}}$

${: \implies \quad \sf d=-3}$

Now , as we got the common difference , so we can find out the missing terms now ;

${: \implies \quad \sf T_{2}= T_{1}+(2-1)d}$

${: \implies \quad \sf T_{2}= 8 +d}$

${: \implies \quad \sf T_{2}= 8-3}$

${: \implies \quad \bf \therefore \: T_{2}= 5}$

Now

${: \implies \quad \sf T_{3}= T_{1}+(3-1)d}$

${: \implies \quad \sf T_{3}= 8 +2d}$

${: \implies \quad \sf T_{3}= 8-6}$

${: \implies \quad \bf \therefore \: T_{3}= 2}$

Also ,

${: \implies \quad \sf T_{4}= T_{1}+(4-1)d}$

${: \implies \quad \sf T_{4}= 8 +3d}$

${: \implies \quad \sf T_{4}= 8-9}$

${: \implies \quad \bf \therefore \: T_{4}= -1}$

Now , The given table can be written as ;

${\begin{array}{|c|c|c|c|c|c|}\cline{1-6} \bf n & \sf 1 & 2 & 3 & 4 & 5 \\ \cline{1-6} \bf T_{n} & \sf 8 & 5 & 2 & -1 & -4 \end{array}}$

Note :- Kindly view the answer from web , if you're not able to see the full answer from here ;

brainly.com/question/26750175

8 0

2 years ago

Repost this for help

inna [77]

Answer:

I'm sorry I don't know if I do get an idea I'll come back and edit

Step-by-step explanation:

3 0

2 years ago

A triangle has one Angle that measures 31° one angle that measures 45° and one angle that measures 140° what kind a triangle is

qaws [65]

unequal Triangle

If two sides of a triangle are unequal, The angle opposite to the greater side is greater than the angle opposite to less.That is,

In a triangle ΔABC,

if AC > AB then ∠B > ∠C

6 0

2 years ago

Solve for the variable x+9=12

labwork [276]

Answer:

Step-by-step explanation:

subtract 9 from both sides

x=3

4 0

2 years ago

Read 2 more answers

A square on a coordinate plane has the points A(0, 0), B(0, 5), C(5, 5) and D(5, 0). If Square ABCD is rotated about the origin

Georgia [21]

Answer:

B. 5 units

Step-by-step explanation:

Transformation is the movement of a point from its initial location to a new location. Types of transformation are reflection, dilation, rotation or translation.

If a point X(x,y) is rotated about the origin 180 degrees clockwise the new point is at X'(-x, -y).

If the square with vertices at A(0, 0), B(0, 5), C(5, 5) and D(5, 0) is rotated about the origin 180 degrees clockwise the new points are at A'(0, 0), B'(0, -5), C'(-5, -5), D'(-5, 0)

A square has four equal sides. The distance between two points $X(x_1,y_1)\ and\ Y(x_2,y_2)$ is:

$|XY|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Hence:

$|A'B'|=\sqrt{(0-0)^2+(-5-0)^2} =5$

3 0

3 years ago