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inessss [21]
3 years ago
10

Angle sum Theory

Mathematics
1 answer:
maria [59]3 years ago
3 0

Step-by-step explanation:

step 1. so ABC is a triangle.

step 2. interior angles of a triangle add up to 180°

step 3. <A + <B + <C = 180°

step 4. 54 + <B + 74 = 180°

step 5. <B = 52°.

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Y∝√x If y=60 when x=36 find,y when x=16
avanturin [10]

Answer:

y = 10√16) = 40

Step-by-step explanation:

This says, "y is proportional to the square root of x."  We are told that y = 60 when x = 36.  Find the formula for this function.  Then, determine what value y will take on when x = 16.

y = k√x  becomes 60 = k√36, or 60 = 6k.  Then k must be 10, and the formula is y = 10√x.

If x = 16, y = 10√16) = 40

7 0
3 years ago
Help I’m so bad at math
gregori [183]

If two triangles are congruent (as per the given congruency statement), then the corresponding angles of both triangles are also congruent.

Therefore, if angle B = 145, then angle E is also 145.

Answer: C) 145

Hope this helps!

8 0
3 years ago
Type the correct answer in the box. Use numerals instead of words. If necessary, use / for the fraction bar.
zimovet [89]
Uhhh ask Siri or get Socratic
5 0
4 years ago
Read 2 more answers
what is the total surface area of the garden in inches, feet and yards that is 70 ft long and 45 wide?
arlik [135]
So,

First, we will find the surface area of the garden in feet.
(70)(45) = 3150 square ft.

Next, to find the surface area of the garden inches, we need to recall the conversion rate between feet and inches.
1 ft. = 12 in.

So, multiply 3150 by 12.
(3150)(12) = 37,800 square in.

To find the surface area of the garden in yards, we need to recall the conversion rate between yards and feet.
1 yd. = 3 ft.

So, divide 3150 by 3.
\frac{3150}{3} = 1050\ square\ yd.

Area of the garden in inches: 37,800 square in.
Area of the garden in feet: 3150 square ft.
Area of the garden in yards: 1050 square yd.
3 0
4 years ago
Solve and list/explain steps, please <br> 2y³ - 9y = -3y²<br><br> College Intermediate Algebra
ss7ja [257]
2y^3 – 2y – 10y + 10 + y^2 – 1 < 0 [the terms are simply reorganized again]
factor 2y from the first two terms, -10 from the second two terms
2y (y^2 -1) – 10 (y-1) + y^2 – 1 < 0
2y (y+1)(y–1) – 10 (y-1) + (y+1)(y–1) < 0 [ because y^2 – 1 = (y+1)(y–1) ]
factor out (y-1) from all the terms
(y-1) [2y(y+1)-10+ y+1] < 0
(y-1) [(y+1) (2y+1) - 10] < 0
Let us simplify (y+1) (2y+1) - 10 < 0 now
(y-1) (2y^2+y+2y+1-10) < 0
(y-1) (2y^2 +3y -9 < 0
(y-1) (2y^2 +6y -3y - 9) < 0 [ because 3y = 6y -3y] j
3 0
3 years ago
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