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Ghella [55]
3 years ago
6

Please help.

Mathematics
1 answer:
Cloud [144]3 years ago
6 0

Answer:

The cups cost $3.00 and the plates cost $4.00

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when running a line, in a right-triangle, from the 90° angle perpendicular to its opposite side, we will end up with three similar triangles, one Small, one Medium and a containing Large one, from where we can use proportions to get their sides, so Check the picture below.

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What is the area of the base in the figure below? A cylinder with height 12 and diameter 6.
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About 28.27.

Step-by-step explanation:

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A = r^2 * pi

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Comparing Linear Functions Written in Different Ways Compare the linear functions expressed below by data in a table and by an e
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Both functions have the same slope.

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8. Name the ray in the figure.<br> B<br> а.<br> ВА<br> b.<br> AB<br> с.<br> ВА<br> d.<br> AB
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3 0
3 years ago
In ΔQRS, s = 2.3 inches, ∠S=51° and ∠Q=44°. Find the area of ΔQRS, to the nearest 10th of an square inch.
postnew [5]

Answer:

Area of ΔQRS = 2.3 square inches

Step-by-step explanation:

From the given information,

<S + <Q + <R = 180^{o}

51 + 44 + <R = 180^{o}

95 + <R = 180^{o}

<R = 180^{o} - 95

    = 85^{o}

<R = 85^{o}

Applying the Sine rule, we have;

\frac{q}{SinQ} = \frac{r}{SinR} = \frac{s}{SinS}

Using \frac{r}{SinR} = \frac{s}{SinS}

\frac{r}{Sin 85} = \frac{2.3}{Sin51}

r = \frac{2.3*Sin85}{sin51}

  = 2.9483

r = 2.9 inches

Also, \frac{q}{SinQ} = \frac{s}{SinS}

\frac{q}{Sin44} = \frac{2.3}{Sin51}

q = \frac{2.3*Sin44}{Sin51}

  = 2.0559

q = 2.0 inches

From Herons formula,

Area of a triangle = \sqrt{s(s-q)(s-r)(s-s)}

s = \frac{2.3 + 2.0 + 2.9}{2}

  = 3.6

Area of ΔQRS = \sqrt{3.6(3.6-2.0(3.6-2.9)(3.6-2.3)}

                        = 2.2895

Area of ΔQRS = 2.3 square inches

8 0
3 years ago
Read 2 more answers
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