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tatuchka [14]
2 years ago
12

Help :(( &&& anyone willing to help me w// algebra I’ll pay

Mathematics
2 answers:
Nookie1986 [14]2 years ago
5 0

Answer:

11

Explanation:

I prefer to get paid in cash but you can pay me the way you want ;)

Aleks04 [339]2 years ago
5 0

Answer:

-21√x + 32

Step-by-step explanation:

Let f(x) = 7x + 4  

g(x) = √x + 4

(-3)f(g(x)) = f (√x + 4)

= -3 [7(√x + 4) + 4]

= -3 [7√x + 28 + 4]

= = -21√x + 32

Even I find this complicated.

Hope its right tho.

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Find the common ratio for the following geometric sequence. 0.75,1.5,3,6,...
Rasek [7]

Answer:

<h2>2</h2>

Step-by-step explanation:

\text{If}\ a_1,\ a_2,\ a_3,\ a_4,\ ...\ ,\ a_n\ \text{is a geometric series, then}\\\\\dfrac{a_2}{a_1}=\dfrac{a_3}{a_2}=\dfrac{a_4}{a_3}=...=\dfrac{a_n}{a_{n-1}}=r=constant

\text{We have}\ 0.75,\ 1.5,\ 3,\ 6,\ ...\\\\\text{The common ratio}\ r:\\\\\dfrac{1.5}{0.75}=2\\\\\dfrac{3}{1.5}=2\\\\\dfrac{6}{3}=2\\\\\boxed{r=2}

4 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
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8w^2=50 i’m not sure how to find the answer
CaHeK987 [17]

8w^2=50

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= w^2= 6.25

=w = √6.25

w= 2.5

8 0
3 years ago
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Vector vector a has a magnitude of 28 units and points in the positive y-direction. When vector vector b is added to vector a ,
Triss [41]

28 + vector b   = -12

vector b  = -40  

vector b has magnitude 40 units in the negative y direction.

5 0
3 years ago
Round 50.057 to the nearest hundredth.
icang [17]
 50.057 = 50.06  <span>to the nearest hundredth.</span>
8 0
3 years ago
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