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ipn [44]
4 years ago
12

How do I do this problem??

Mathematics
1 answer:
Montano1993 [528]4 years ago
8 0
The answer to this problem is 1 :)
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The coordinates of a triangle's vertices are A(2, 4), B(3, 1), and C(5, 5). The triangle undergoes a translation 4 units to the
AlekseyPX
South west

Shifting 4 units left (decreasing in x-axis by 4 units)
Shifting 5 units down (decreasing in y-axis by 5 units)

Apply to every vertices coordinates(x - 4, y - 5)

New coordinates

A(-2, -1)
B(-1, -4)
C(1,0)
5 0
3 years ago
In the figure above, PQRS is a circle. If PQT and SRT<br>are straight lines, find the value of x.
Elena L [17]

Given:

PQRS is a circle, PQT and SRT  are straight lines.

To find:

The value of x.

Solution:

Since PQRS is a circle, PQT and SRT  are straight lines, therefore, PQRS isa cyclic quadrilateral.

We know that, sum of opposite angles of a cyclic quadrilateral is 180 degrees.

m\angle SPQ+m\angle QRS=180^\circ

81^\circ+m\angle QRS=180^\circ

m\angle QRS=180^\circ-81^\circ

m\angle QRS=99^\circ

Now, SRT  is a straight line.

m\angle QRT+m\angle QRS=180^\circ             (Linear pair)

m\angle QRT+99^\circ=180^\circ

m\angle QRT=180^\circ-99^\circ

m\angle QRT=81^\circ               ...(i)

According to the Exterior angle theorem, in a triangle the measure of an exterior angle is equal the sum of the opposite interior angles.

Using exterior angle theorem in triangle QRT, we get

m\angle PQR=m\angle QRT+m\angle QTR

x=81^\circ+22^\circ

x=103^\circ

Therefore, the value of x is 103 degrees.

4 0
3 years ago
A person is standing on a cliff that is 200ft above a body of water. The person looks down at an angle of depression of 42o at a
velikii [3]

Answer:

524.5 feet

Step-by-step explanation:

Given: Height of cliff= 200 ft

           Angle of depression at sail boat is 42°

           Angle of depression at yacht is 15°

Lets assume distance sailboat from the bottom of cliff is "d_1"

And assume distance yacht from the bottom of cliff is "d_2"

Now using tangent rule to solve it.

we know, tan\theta = \frac{opposite}{adjacent}

Distance sailboat from the bottom of cliff; tan 42= \frac{200}{d_1}

Using trignometry table to know the value of tan 42°

⇒0.90= \frac{200}{d_1}

cross multiplying both side.

⇒ d_1= \frac{200}{0.90} = 222.22 \approx 222

∴ Distance sailboat from the bottom of cliff (d_1)= 222 feet

Distance yatch from the bottom of cliff; tan 15= \frac{200}{d_2}

Using trignometry table to know the value of tan 15°

⇒0.2679= \frac{200}{d_2}

cross multiplying both side.

⇒ d_2= \frac{200}{0.2679} = 746.54 \approx 746.5

∴Distance yacht from the bottom of cliff (d_2)= 746.5 feet.

Next, finding the distance between sailboat and yacht.

Distance between sailboat and yacht = d_2-d_1

⇒ Distance between sailboat and yacht= 746.5-222= 524.5\ ft

∴ Distance between sailboat and yacht is 524.5 feet.

3 0
3 years ago
NEED HELP ON THIS PLEASEEEEEEEEEEE
solong [7]

Answer: i am a 7th grade child and i dont know an idea of how to solve this..... just kidding but its b :) hopefully this will help u

Step-by-step explanation:

5 0
3 years ago
Find the area of the trapezoid with bases 12 cm and 14 cm and height 9 cm
Cerrena [4.2K]

Answer:

117

Step-by-step explanation:

b1+b2÷2×h is the formula.

Just fill in the formula with the numbers you have.

12+14=26

26÷2=13

13×9=117

3 0
3 years ago
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