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goblinko [34]
3 years ago
13

Round 27.68 to the nerest tenth​

Mathematics
2 answers:
aleksandr82 [10.1K]3 years ago
7 0

Answer:

27.7

Step-by-step explanation:

Helen [10]3 years ago
3 0

Answer:

28

Step-by-step explanation:

68 is closer to hundred

or you can say 27.70

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Help please explain as simple as possible
Virty [35]
Short answer: (-8)^2 + 8  x  -8 = 0

Use PEMDAS

"Evaluate the expression" just means solve until you can't simplify anymore. You must solve it in a certain order according to PEMDAS: Parentheses, Exponents, Multiply, Divide, Add, Subtract.

What does the beginning of the expression look like? It is (-8)^{2}.

According to PEMDAS, you must solve what is in the parentheses *first*. But, since there is only a number (-8), there is nothing to solve for and you can move on to exponents.

The squared symbol, the little 2, means you have to square what is *inside* the parentheses. (-8)^{2} = 64, because -8 times itself is 64.

Next comes multiplication. Remember, we are not working from left to right. We must multiply the values on the far right before we do any adding, because multiplication comes *before* addition.

(64) + (8 times -8)

(64) + (-64)

Finally, we can add. In this case, because we are adding a negative number, we are really subtracting. 64 + -64 equals 0.
8 0
3 years ago
For the parallelogram ABCD the extensions of the angle bisectors AG and BH intersect at point P. Find the area of the parallelog
natita [175]

Answer:

The area of a parallelogram is 360 in.²

Step-by-step explanation:

Where DG = GH

GP = 12 in.

AB = 39 in.

∠DAB + ∠ABC = 180° (Adjacent angles of a parallelogram)

Whereby ∠DAB is bisected by AG and ∠ABC is bisected by BH

Therefore, ∠GAB + ∠HBA = 90°

Hence, ∠BPA = 90° (Sum of interior angles of a triangle)

cos(\angle GAB) = \dfrac{AP}{AB} = \dfrac{AP}{39} = \dfrac{GP}{GH} =\dfrac{12}{GH}

We note that ∠AGD = ∠GAB (Alternate angles of parallel lines)

∴ ∠AGD = ∠AGD since ∠AGD = ∠GAB (Bisected angle)

Hence AD = DG (Side length of isosceles triangle)

The bisector of ∠ADG is parallel to BH and will bisect AG at point Q

Hence ΔDAQ  ≅ ΔDGQ ≅ ΔGPH and AQ = QG = GP

Hence, AP = 3 × GP = 3 × 12 = 36

cos(\angle GAB) = \dfrac{AP}{AB} = \dfrac{36}{39}

\angle GAB = cos^{-1} \left (\dfrac{36}{39}  \right )

∠GAB = 22.62°

cos(\angle GAB) =  \dfrac{36}{39} = \dfrac{12}{GH}

GH =  \dfrac{39}{36} \times {12}

GH = 13 in.

∴ AD 13 in.

BP = 39 × sin(22.62°) = 15 in.

GH = √(GP² + HP²)

∠DAB = 2 × 22.62° = 45.24°

The height of the parallelogram = AD × sin(∠DAB) =  13 × sin(45.24°)

The height of the parallelogram = 120/13 =  9.23 in.

The area of a parallelogram = Base × Height = (120/13) × 39 = 360 in.²

7 0
3 years ago
Read 2 more answers
Date
ss7ja [257]

Answer:

hshshxhdjebrbhdjwjdbevvdhdhfiiehrvrvdhdjrirubrvfgdhjdjejrhvrvdghfueyrggdvdvdgdhdudhdhegdghdhhrhdbdbdbhdhdhfhfydhhdbfbdbdbrbrhhrhrhrhfhfuyfh4h4bbdbdhxufhfbbdhdhdhrhffbfbdbduc

7 0
3 years ago
What's the expanded form to write 16,107,320
Romashka [77]
(10,000,000) + (6,000,000) + (100,000) + (7,000) + (300) + (20)
6 0
3 years ago
How do i solve the question in the image provided ?
Mars2501 [29]

Answer:

0.19+0.46i and 0.19-0.46i

Step-by-step explanation:

\neq (-b+\sqrt{b^2-4ac} )/2a\\and\\(-b-\sqrt{b^2-4ac} )/2a\\\\a=8\\b=3\\c=2\\(-3+\sqrt{3^2-4*8*2} )/(2*8)\\= - 0.1875+0.463512i\\(-3-\sqrt{3^2-4*8*2} )/(2*8) = -0.1875-0.463512i

hope its right

6 0
3 years ago
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