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mojhsa [17]
3 years ago
8

Help please I’ve been looking online for help but couldn’t find any.

Mathematics
1 answer:
alexandr1967 [171]3 years ago
4 0
5 is 33 because the question ends up being 14+3+16
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Find the Area of the Trapezoid.<br><br> A. 7.5 ft^2<br> B. 9ft ^2<br> C. 10.5 ft^2<br> D. 18 ft^2
erica [24]

Answer:

B

Step-by-step explanation:

8 0
3 years ago
3rd question, I’m sorry guys but I was sick that’s why I didn’t attend school and the teacher said if I didn’t submit it by 5 ho
aniked [119]

Answer:

21.  ΔUYZ <u>Obtuse</u>

22. ΔBCD <u>Right or Rectangle</u>

23. ΔADB <u>Acute</u>

24. ΔUXZ <u>Acute</u>

25. ΔUWZ <u>Right or Rectangle</u>

26. ΔUXY <u>Equiangular and acute</u>

Step-by-step explanation:

We need to classify each triangle by its angles.

Acute: They have 3 acute angles (less than 90 degrees).

Rectangle: The inner angle A is straight (90 degrees) and the other 2 angles are sharp. The sides that form the right angle are called legs (c and b), the other side hypotenuse.

Obtuse: The inner angle A is obtuse (more than 90 degrees). The other 2 angles are acute.

ΔUYZ Obtuse  The inner angle Y is obtuse (more than 90°).

ΔBCD Right or Rectangle The inner angle C is straight (90°).

ΔADB Acute It has 3 acute angles (less than 90°).

ΔUXZ Acute  It has 3 acute angles (less than 90°).

ΔUWZ Right or Rectangle The inner angle W is straight (90°).

ΔUXY Equiangular and acute Which means all the inner angles are equal (60°).

3 0
3 years ago
PLEASE HELP IM DESPERATE precalculus. also 50 points!! please god help
kompoz [17]

Answer:

For B thru F these options will vary but here how you do it

B. Step 1 Draw the 4 Quadrants.

Then Draw the Triangle in the lower right quadrant which we call quadrant 4. Label the X axis as Adjacent and positive. Label the Y axis as Opposite and negative. Label the Slanted side as the hypotune and AS POSITVE SINCE HYPOTENUSE IS ALWAYS POSITIVE.

FOR C. IN QUADRANT 2, PLOT A POINT AT 0,12 AND AT (-5,0). CONNECT THE DOTS AND IT FORMS A TRIANGLE. Label the X axis as adjacent and negative and the y axis as positve and opposite and label the slanted side hypotunese and positive.

FOR D Draw a straight line along the x axis then draw a slanted line passing through (5,-1). In between them put the theta symbol in there.

The labeling is the same for C.

For E. Since tan must be positve and secant must be positve, our triangle must be in the 1st Quadrant. Draw any right triangle as long it is in the first quadrant

The x axis is adjacent and positve. The y axis is opposite and positve. The hypotenuse is the slanted side and it is positve.

For F. Since sin is negative and cos is positve the triangle is in the 4th quadrant. Draw any triangle in the 4th quadrant and the labeling is the same for Problem B.

2. We can find the sec of cos by flipping cosine.

\cos( \frac{x}{y} )  =  \sec( \frac{y}{x} )

\cos( \frac{1}{2} )  =  \sec(2 )

Sec is 2.

To find the cotangent, first let find the sin then tan.

We can use the identity

\cos( {theta}^{2} )  +  \sin( {theta}^{2} )  = 1

Let plug in the number

\cos( \frac{ {1}^{2} }{{2}^{2} } )  +  \sin(x {}^{2} )  = 1

\cos( \frac{1}{4} )  +  \sin(x {}^{2} )  = 1

\sin(x {}^{2} )  =   1 -  \frac{1}{4}

\sin(x {}^{2} )  =  \frac{3}{4}

\sin(x)  =  \frac{ \sqrt{3} }{ \sqrt{4} }

\sin(x)  =  \frac{ \sqrt{3} }{2}

Since sin is negative, sin x=

-  \frac{ \sqrt{3} }{2}

Now let apply the formula

\frac{ \sin(x) }{ \cos(x) }  =  \tan(x)

\frac{ \frac{ -  \sqrt{3} }{2} }{ \frac{1}{2} }  =  \tan(x)

-  \sqrt{3}

Now let find cotangent we can the reciprocal of

tan.

\tan=  -  \sqrt{3}

\cot =   - \frac{1}{ \sqrt{3} }

Rationalize denominator

\frac{ - 1}{ \sqrt{3} }  \times   \frac{ \sqrt{3} }{ \sqrt{3} }  =  \frac{ \sqrt -{ 3} }{3}

cotangent equal

-  \frac{ \sqrt{ 3} }{3}

4 0
3 years ago
Please please help it is very urgent!​
Liula [17]
Y = -0.33x + -5.00 (rounded to the nearest 100)
7 0
3 years ago
Can u please help me
Zina [86]

Option A would be the best strategy here.

4 0
3 years ago
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