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1 year ago

5

Jake shows that △ △ C B A is congruent to △ △ A D C by rotating △ 180° △ C B A 180 ° around point C so it matches up with △ △ A

D C exactly. Which conclusion can be drawn from Jake's transformations?

1 answer:

Naddik [55]1 year ago

5 0

Answer:

Reflection about either the x- axis or y- axis.

Step-by-step explanation:

It is likely that the triangle shows a mirror image when it has been rotated at point C so that it matches ΔADC.

The triangles are congruent if the sides and angles are equal. The only way to effect the change is through a rotation.

In fact, two objects are said to be congruent if they can be transformed into another shape by translation, rotation, and reflection.

The resulting image is a distinct figure that can fit and match the other completely.

In this case, the triangle is rotated through 180⁰ so the resulting image is a reflection or mirror image.

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Answer:

The answer is the option A

$96\ in^{2}$

Step-by-step explanation:

we know that

The lateral area of a right rectangular prism is equal to

$LA=P*h$

where

P is the perimeter of the base

h is the height of the prism

Find the perimeter

$P=2*(3+5)=16\ in$

$h=6\ in$

substitute in the formula

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1 year ago

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Bella has joined a new gym in town. The cost of membership is $25 per month. The after-hours policy at the gym allows her to wor

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The total monthly bill of the gym = $53

The cost of membership of a month = $25

Let 'n' be extra the number of hours Bella worked on.

The cost for working on extra hours = $4

So, we have to determine the equation, Bella worked out after hours.

We will determine the equation by:

(Monthly cost of membership) + ( cost for extra hours $\times$ number of hours extra worked on ) = Total monthly bill received

So, we get

$\$25+(4 \times n) = \$53$

$25+4n = $53 is the required equation.

Therefore, $25+4n = $53 equation can be used to determine how many times Bella worked out after hours.

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1 year ago

On a december day, the probability of snow is .30. the probability of a "cold" day is .50. the probability of snow and "cold" we

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Events A and B are independent if the following holds true:

P(A ∩ B) = P(A) * P(B)

where P(A ∩ B) is the probability of A and B, P(A) is the probability of A, and P(B) is the probability of B.

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2 years ago

Please answer all of them need this

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First Question

For a better understanding of the solution provided here please find the first attached file which has the diagram of the the isosceles trapezoid.

We dropped perpendiculars from C and D to intersect AB at Q and P respectively.

As can be seen in $\Delta BCQ$ , we can easily find the values of CQ and BQ.

Since, $Sin(75^0)=\frac{CQ}{8}$

$\therefore CQ=8\times Sin(75^0)\approx 7.73$ ft

In a similar manner we can find BQ as:

$Cos(75^0)=\frac{BQ}{8}$

$BQ\approx2.07$ ft

All these values can be found in the diagram attached.

Thus, because of the inherent symmetry of the isosceles trapezoid, PQ can be found as:

$PQ=22-(AP+QB)=22-(2.07+2.07)=17.86$

Let us now consider $\Delta AQC$

We can apply the Pythagorean Theorem here to find the length of the diagonal AC which is the hypotenuse of $\Delta AQC$ .

$AC=\sqrt{(AQ)^2+(QC)^2}=\sqrt{(AP+PQ)^2+(QC)^2}=\sqrt{(2.07+17.86)^2+(7.73)^2}\approx21.38$ feet.

Thus, out of the given options, Option B is the closest and hence is the answer.

Second Question

For this question we can directly apply the formula for the area of a triangle using sines which is as:

Area= $\frac{1}{2}$ (First Side)(Second Side)(Sine of the angle between the two sides)

Thus, from the given data,

Area= $\frac{1}{2}\times 218.5\times 224.5\times sin(58.2^0)\approx20845$ $m^2$

Therefore, Option D is the correct option.

Third Question

For this question we will apply the Sine Rule to the $\Delta ABC$ given to us.

Thus, from the triangle we will have:

$\frac{AB}{Sin(\angle C)}=\frac{BC}{Sin(\angle A)}$

$\frac{c}{Sin(\angle C)}=\frac{a}{Sin(\angle A)}$

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This gives $a$ to be:

$a\approx28.44$

Which is not close to any of the given options.

Fourth Question

Please find the second attachment for a better understanding of the solution provided her.

As can be clearly seen from the attached diagram, we can apply the Cosine Rule here to find the return distance of the plane which is CA.

$AC=\sqrt{(AB)^2+(BC)^2-2(AB)(BC)\times Cos(\angle B)}$

$\therefore AC=\sqrt{(172.20)^2+(111.64)^2-2(172.20)(111.64)\times Cos(177.29^0)}\approx283.8$ miles.

Thus, Option D is the answer.

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