Use the order of operations to simplify this expression. 2 + 4(5 − 8)

A right triangle with coordinates A (－1, 4), B(－1, 8) and C (3, 4) is first rotated 90 degrees counterclockwise and then transla

Sophie [7]

Answer:

The ∠B prime A prime C prime in the resulting figure = 90°

Step-by-step explanation:

The coordinates of the right triangle = A( -1, 4), B(-1, 8) and C(3, 4)

The rotation 90° counterclockwise of a point on a preimage (x, y) gives the coordinates of the location of the point of the image after rotation as (-y, x)

Therefore we have;

Applying 90° counterclockwise rotation we have;

The point A(-1, 4) is relocated to the point (-4, -1)
The point B(-1, 8) is relocated to the point (-8, -1)
The point C(3, 4) is relocated to the point (-4, 3)

Applying a translation T(-2, -3) to the new points above which is a translation 2 units left and 3 units down, we have;

The coordinates of the point A prime is (-4 - 2, -1 - 3) = (-6, -4)

The coordinates of the point B prime is (-8 - 2, -1 - 3) = (-10, -4)

The coordinates of the point C prime is (-4 - 2, 3 - 3) = (-6, 0)

The coordinates of the vertices of the triangle ΔA prime B prime C prime are A prime (-6, -4) B prime (-10, -4) C prime (-6, 0)

The measure of the angle ∠B prime A prime C prime is given as follows;

Length of a segment of each segment of the triangle are found using the following equation;

$l = \sqrt{\left (x_{2}-x_{1} \right )^{2}+\left (y_{2}-y_{1} \right )^{2}}$

Which gives;

Length of B prime A prime = √(((-10) - (-6))² + ((-4) - (-4))²) = 4

Length of B prime C prime = √(((-10) - (-6))² + ((-4) - 0)²) = 32 = 4·√2

Length of C prime A prime = √(((-6) - (-6))² + ((0) - (-4))²) = 4

∴ B prime C prime is the hypotenuse side and the ∠B prime A prime C prime = The angle opposite to the hypotenuse side = 90°

The ∠B prime A prime C prime in the resulting figure = 90°

6 0

3 years ago

In right triangle ΔABC (m∠C = 90°), point P is the intersection of the angle bisectors of the acute angles. The distance from P

arlik [135]

Answer:

28 inches

Step-by-step explanation:

The point of intersection of the angle bisectors is the incenter. It is the center of a circle tangent to the three sides of the triangle. The circle has radius 2.

In the attached figure, we have labeled the points of tangency D, E, and F. We know that CE and CF are both of length 2, and we know that the points of tangency are the same distance from an external point where the tangents intersect. That means DA = FA and DB = EB.

The perimeter of the triangle is ...

P = DA +DB +FA +EB +CF +CE

Using the above relations, this can be written as ...

P = DA +DB +DA +DB +CF +CE = 2(DA +DB) +2(CE)

We are told that AB is 12 inches, so DA +DB = 12 inches. We also know that CE = 2 inches, so the perimeter is ...

P = 2(12 in) + 2(2 in) = 28 in

The perimeter of triangle ABC is 28 inches.

4 0

3 years ago

Researchers once surveyed students on which superpower they would most like to have. The following two-way table displays data f

GenaCL600 [577]

Answer:

The percent of students in the sample who were male is 48%.

Step-by-step explanation:

The complete question is:

Researchers once surveyed 100 students on which superpower they would most like to have. The following two-way table displays data for the sample of students who responded to the survey.

Superpower Male Female TOTAL

Fly 26 12 38

Invisibility 12 32 44

Other 10 8 18

TOTAL 48 52 100

What percent of students in the sample were male?

Solution:

The probability of an event E is the ratio of the number of favorable outcomes to the total number of outcomes.

$P(E)=\frac{n(E)}{N}$

The number of males in the sample is, n (M) = 48.

Total number of students, N = 100.

Compute the probability of selecting a male student as follows:

$P(M)=\frac{n(M)}{N}\\$

$=\frac{48}{100}\\\\=0.48$

Then the percent of students in the sample who were male is:

$\text{Percent of Males}=P(M)\times 100\%$

$=0.48\times 100\%\\\\=48\%$

Thus, the percent of students in the sample who were male is 48%.

4 0

3 years ago

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Exploring Slopes of Parallel and Perpendicular Lines - Item 32539

bekas [8.4K]

What’s the picture of problem

6 0

3 years ago

Write one digit on the both sides of 57 to make the number divisible by 72. How many solutions does this problem have?

Brilliant_brown [7]

I'm assuming writing "one digit on the both sides of 57" means you write the same digit to either side, like in 1571?

Given a number $n$ with $k$ digits, prepending and appending the same digit $d$ , $1\le d\le9$ (omit 0 because it doesn't change the starting number), is the same as multiplying $n$ by 10 and adding $(10^{k+1}+1)d$ . We have $n=57$ , so we're looking for $d$ such that

$570+(10^3+1)d=1001d+570\equiv0\pmod{72}$

8 and 9 are coprime, so we can use the Chinese remainder theorem.

$72=8\cdot9\implies\begin{cases}1001d+570\equiv d+2\equiv0\pmod8\\1001d+570\equiv2d+3\equiv0\pmod9\end{cases}$

which simplifies to

$\begin{cases}d\equiv6\pmod8\\d\equiv3\pmod9\end{cases}$

Now we apply the CRT. We want some number $d$ such that, taken modulo 8, returns a remainder of 6, but taken modulo 9, returns a remainder of 3. So we could try

$d=6\cdot9+3\times8=54+24=78$

Modulo 8, the second term vanishes, and $54\equiv6\pmod8$ . However, modulo 9, the first term vanishes but $24\equiv6\equiv2\cdot3\pmod9$ , whereas we only want 3. So we multiply the second term by the inverse of 2 modulo 9.

To find the inverse, notice that $10\equiv2\times5\equiv1\pmod9$ , so $2^{-1}\equiv5\pmod9$ , and so we multiply the second term by 5. Now,

$d=6\cdot9+3\times8\times5=174$

and

$174\equiv72\cdot2+30\equiv30\pmod{72}$

so we find that $d=72n+30$ . In particular, the smallest positive solution is 30, which is larger than 9 so there are no single-digit choices of $d$ that makes the new number divisible by 72.

4 0

3 years ago

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