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attashe74 [19]
4 years ago
15

Omg my daughter and I are totally lost on this homework. Please help. It is multiplying decimals. She has to turn this in today.

Mathematics
1 answer:
Wewaii [24]4 years ago
5 0

Answer:

5 x 0.6 = 5 x 6tenths, as 6/10 = 0.6

= 30 tenths

7 x 0.8 = 7 x 8tenths, as 8/10 is 0.8

= 56 tenths

Step-by-step explanation:

If you really want a more detailed explanation, just reply to this message.

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What is the unit form using the largest unit possible for 4tens+6tens
Gre4nikov [31]
40 + 60 = 100

Hundrends place
4 0
3 years ago
Please solve with explanation (high points)
Hunter-Best [27]

Step-by-step explanation:

so, we have a large triangle made of the 2 cables as legs and the ground distance AB as baseline.

the tower is the height to the baseline of that large triangle.

let's call the top of the tower T.

and remember, the sum of all angles in a triangle is always 180°.

we know the angle A = 62°, and angle B = 72°.

assuming that AB is a truly horizontal line that means that the 2 legs (cables) have different lengths, the triangle is not isoceles, and the tower is not in the middle of the baseline.

so, the height (tower) splits the baseline into 2 parts. let's call them p and q.

p + q = 12 m

p = 12 - q

let's simply define that p is the part of the baseline on the A side, and q is the part of the baseline on the B side.

we have now 2 small right-angled triangles the large height (tower) splits the large triangle into.

one has the sides

AT, height (tower), p

angle A = 62°

angle T = 180 - 90 - 62 = 28°

the other has the sides

BT, height (tower), q

angle B = 72°

angle T = 180 - 90 - 72 = 18°

now remember the law of sine :

a/sin(A) = b/sin(B) = c/sin(C)

with the sides and the associated angles being opposite.

p/sin(28) = height/sin(62)

q/sin(18) = height/sin(72)

we know from above that

p = 12 - q

so,

(12 - q)/sin(28) = height/sin(62)

height = (12 - q)×sin(62)/sin(28)

q/sin(18) = height/sin(72)

height = q×sin(72)/sin(18)

and therefore, as height = height we get

(12 - q)×sin(62)/sin(28) = q×sin(72)/sin(18)

(12 - q)×sin(62)×sin(18) = q×sin(72)×sin(28)

12×sin(62)×sin(18) - q×sin(62)×sin(18) =

= q×sin(72)×sin(28)

12×sin(62)×sin(18) = q×sin(72)×sin(28) + q×sin(62)×sin(18) =

= q×(sin(72)×sin(28) + sin(62)×sin(18))

q = 12×sin(62)×sin(18) / (sin(72)×sin(28) + sin(62)×sin(18))

q = 4.551603755... m

p = 12 - q = 7.448396245... m

height = q×sin(72)/sin(18) = 14.00839594... m ≈ 14 m

the cell tower is about 14 m tall.

7 0
2 years ago
Hey loves, plz help if you have time:)
aksik [14]

Answer:

ΔABX and ΔEDX

Step-by-step explanation:

Given: AX ≅ EX

          BX ≅ DX

SAS implies a congruent relation with respect to Side-Angle-Side between the two required triangle.

AX ≅ EX (given)

BX ≅ DX (given)

<BXA ≅ <DXE (vertical opposite angles)

Therefore,

ΔABX ≅ ΔEDX (Side-Angle-Side, SAS, congruence property)

3 0
3 years ago
Find the domain/range and equation
vichka [17]

Answer:

Domain = [1.5, 4.5]

Range  = [4, ∞)

Step-by-step explanation:

The domain of the graph are the values along the input variable x while the range are the values of the output variable y;

From the figure the domain values falls with the range [-1.5, -4.5] while the range are with thin the values [4, ∞)

6 0
3 years ago
The points A(1, 4), B(5,1) lie on a circle. The line segment AB is a chord. Find the equation of a diameter of the circle.
tangare [24]

Check the picture below.

well, we want only the equation of the diametrical line, now, the diameter can touch the chord at any several angles, as well at a right-angle.

bearing in mind that <u>perpendicular lines have negative reciprocal</u> slopes, hmm let's find firstly the slope of AB, and the negative reciprocal of that will be the slope of the diameter, that is passing through the midpoint of AB.

\bf A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{4}}}{\underset{run} {\underset{x_2}{5}-\underset{x_1}{1}}}\implies \cfrac{-3}{4} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{slope of AB}}{-\cfrac{3}{4}}\qquad \qquad \qquad \stackrel{\textit{\underline{negative reciprocal} and slope of the diameter}}{\cfrac{4}{3}}

so, it passes through the midpoint of AB,

\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{5+1}{2}~~,~~\cfrac{1+4}{2} \right)\implies \left(3~~,~~\cfrac{5}{2} \right)

so, we're really looking for the equation of a line whose slope is 4/3 and runs through (3 , 5/2)

\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{\frac{5}{2}}) \stackrel{slope}{m}\implies \cfrac{4}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{\cfrac{5}{2}}=\stackrel{m}{\cfrac{4}{3}}(x-\stackrel{x_1}{3})\implies y-\cfrac{5}{2}=\cfrac{4}{3}x-4 \\\\\\ y=\cfrac{4}{3}x-4+\cfrac{5}{2}\implies y=\cfrac{4}{3}x-\cfrac{3}{2}

4 0
3 years ago
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