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2 years ago

Assume that TUV=WXY which of the following congruence statements are correct

Mathematics

2 answers:

Damm [24]2 years ago

7 0

Answer:

A and E are correct option.

Step-by-step explanation:

We are given two triangle congruent.

If two triangles are congruent then their corresponding sides and angles are equal.

In ΔTUV ≅ ΔWXY

Now we write all the congruent sides and angles

∠T=∠W
∠U=∠X
∠V=∠Y
TU≅WX
UV≅XY
TV≅WY

Now we see all the given option.

∠Y=∠V (TRUE, Congruent part of congruence triangles)

∠W=∠T (TRUE, Congruent part of congruence triangles)

Thus, A and E are correct option.

Assoli18 [71]2 years ago

4 0

Your answers should be A and E

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Zöe schedules advertising for a radio station. She must fill 12 minutes each hour with 30 second ads and 60 second ads. Zöe sold

PilotLPTM [1.2K]

I cannot see Zoe's work to explain the error, but the correct method of solving is listed:

x is the number of 30-second ads
y is the number of 60-second ads

x+y=12(60)=720 would be the first equation; this is because while the ads together make 12 minutes, the ad times are in seconds. This means we must multiply 12 by 60.

y=2x is the second equation

Our system is then
x+y=720
y=2x

We will use substitution to solve this. Plug 2x in place of y in the first equation:
x+2x = 720

Combine like terms:
3x = 720

Divide both sides by 3:
3x/3 = 720/3
x = 240

Substitute this value in for x in the second equation:
y=2(240)
y=480

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2 years ago

A car was purchased for $25,000. Research shows that the car has an average yearly depreciation rate of 18.5%. Create a function

vitfil [10]

Answer:

V(t) = 25000 * (0.815)^t

The depreciation from year 3 to year 4 was $2503.71

Step-by-step explanation:

We can model V(t) as an exponencial function:

V(t) = Vo * (1+r)^t

Where Vo is the inicial value of the car, r is the depreciation rate and t is the amount of years.

We have that Vo = 25000, r = -18.5% = -0.185, so:

V(t) = 25000 * (1-0.185)^t

V(t) = 25000 * (0.815)^t

In year 3, we have:

V(3) = 25000 * (0.815)^3 = 13533.58

In year 4, we have:

V(4) = 25000 * (0.815)^4 = 11029.87

The depreciation from year 3 to year 4 was:

V(3) - V(4) = 13533.58 - 11029.87 = $2503.71

7 0

2 years ago

The center of a circle is at the origin on a coordinate grid. The vertex of a parabola that opens upward is at (0, 9). If the ci

zhannawk [14.2K]

Answer:

"The maximum number of solutions is one."

Step-by-step explanation:

Hopefully the drawing helps visualize the problem.

The circle has a radius of 9 because the vertex is 9 units above the center of the circle.

The circle the parabola intersect only once and cannot intercept more than once.

The solution is "The maximum number of solutions is one."

Let's see if we can find an algebraic way:

The equation for the circle given as we know from the problem without further analysis is so far $x^2+y^2=r^2$ .

The equation for the parabola without further analysis is $y=ax^2+9$ .

We are going to plug $ax^2+9$ into $x^2+y^2=r^2$ for $y$ .

$x^2+y^2=r^2$

$x^2+(ax^2+9)^2=r^2$

To expand $(ax^2+9)^2$ , I'm going to use the following formula:

$(u+v)^2=u^2+2uv+v^2$ .

$(ax^2+9)^2=a^2x^4+18ax^2+81$ .

$x^2+y^2=r^2$

$x^2+(ax^2+9)^2=r^2$

$x^2+a^2x^4+18ax^2+81=r^2$

So this is a quadratic in terms of $x^2$

Let's put everything to one side.

Subtract $r^2$ on both sides.

$x^2+a^2x^4+18ax^2+81-r^2=0$

Reorder in standard form in terms of x:

$a^2x^4+(18a+1)x^2+(81-r^2)=0$

The discriminant of the left hand side will tell us how many solutions we will have to the equation in terms of $x^2$ .

The discriminant is $B^2-4AC$ .

If you compare our equation to $Au^2+Bu+C$ , you should determine $A=a^2$

$B=(18a+1)$

$C=(81-r^2)$

The discriminant is

$B^2-4AC$

$(18a+1)^2-4(a^2)(81-r^2)$

Multiply the (18a+1)^2 out using the formula I mentioned earlier which was:

$(u+v)^2=u^2+2uv+v^2$

$(324a^2+36a+1)-4a^2(81-r^2)$

Distribute the 4a^2 to the terms in the ( ) next to it:

$324a^2+36a+1-324a^2+4a^2r^2$

$36a+1+4a^2r^2$

We know that $a>0$ because the parabola is open up.

We know that $r>0$ because in order it to be a circle a radius has to exist.

So our discriminat is positive which means we have two solutions for $x^2$ .

But how many do we have for just $x$ .

We have to go further to see.

So the quadratic formula is:

$\frac{-B \pm \sqrt{B^2-4AC}}{2A}$

We already have $B^2-4AC}$

$\frac{-(18a+1) \pm \sqrt{36a+1+4a^2r^2}}{2a^2}$

This is t he solution for $x^2$ .

To find $x$ we must square root both sides.

$x=\pm \sqrt{\frac{-(18a+1) \pm \sqrt{36a+1+4a^2r^2}}{2a^2}}$

So there is only that one real solution (it actually includes 2 because of the plus or minus outside) here for x since the other one is square root of a negative number.

That is,

$x=\pm \sqrt{\frac{-(18a+1) \pm \sqrt{36a+1+4a^2r^2}}{2a^2}}$

means you have:

$x=\pm \sqrt{\frac{-(18a+1)+\sqrt{36a+1+4a^2r^2}}{2a^2}}$

$x=\pm \sqrt{\frac{-(18a+1)-\sqrt{36a+1+4a^2r^2}}{2a^2}}$ .

The second one is definitely includes a negative result in the square root.

18a+1 is positive since a is positive so -(18a+1) is negative

2a^2 is positive (a is not 0).

So you have (negative number-positive number)/positive which is a negative since the top is negative and you are dividing by a positive.

We have confirmed are max of one solution algebraically. (It is definitely not 3 solutions.)

If r=9, then there is one solution.

If r>9, then there is two solutions as this shows:

$x=\pm \sqrt{\frac{-(18a+1)+\sqrt{36a+1+4a^2r^2}}{2a^2}}$

r=9 since our circle intersects the parabola at (0,9).

Also if (0,9) is intersection, then

$0^2+9^2=r^2$ which implies r=9.

Plugging in 9 for r we get:

$x=\pm \sqrt{\frac{-(18a+1)+\sqrt{36a+1+4a^2(9)^2}}{2a^2}}$

$x=\pm \sqrt{\frac{-(18a+1)+\sqrt{36a+1+324a^2}}{2a^2}}$

$x=\pm \sqrt{\frac{-(18a+1)+\sqrt{(18a+1)^2}}{2a^2}}$

$x=\pm \sqrt{\frac{-(18a+1)+18a+1}{2a^2}}$

$x=\pm \sqrt{\frac{0}{2a^2}}$

$x=\pm 0$

$x=0$

The equations intersect at x=0. Plugging into $y=ax^2+9$ we do get $y=a(0)^2+9=9$ .

After this confirmation it would be interesting to see what happens with assume algebraically the solution should be (0,9).

This means we should have got x=0.

$0=\frac{-(18a+1)+\sqrt{36a+1+4a^2r^2}}{2a^2}$

A fraction is only 0 when it's top is 0.

$0=-(18a+1)+\sqrt{36a+1+4a^2r^2}$

Add 18a+1 on both sides:

$18a+1=\sqrt{36a+1+4a^2r^2$

Square both sides:

$324a^2+36a+1=36a+1+4a^2r^2$

Subtract 36a and 1 on both sides:

$324a^2=4a^2r^2$

Divide both sides by $4a^2$ :

$81=r^2$

Square root both sides:

$9=r$

The radius is 9 as we stated earlier.

Let's go through the radius choices.

If the radius of the circle with center (0,0) is less than 9 then the circle wouldn't intersect the parabola. So It definitely couldn't be the last two choices.

7 0

2 years ago

Read 2 more answers

Translate triangle a by vector (-3/1)to give triangle b. Then rotate your triangle b 180 degrees around the origin to give trian

Dmitrij [34]

Answer:

Rotation of triangle A across a point (1.5, -0.5) by 180°.

Step-by-step explanation:

Translation of a figure about a vector $\binom{-3}{1}$ means translation of a figure by 3 units to the left on x-axis and 1 unit in the positive direction of the y-axis.

Rule for the translation will be,

(x, y) → [(x - 3), (y + 1)]

From the figure attached,

Vertices of the triangle A are A(1, -1), B(1, -4) and C(3, -4)

After translation new image points will be A'(-2, 0), B'(-2, -3), C'(0, -3).

After rotation of these image points 180° about the origin will be,

Rule for the rotation,

(x, y) → (-x, -y)

A'(-2, 0) → A"(2, 0)

B'(-2, -3) → B"(2, 3)

C'(0, -3) → C"(0, 3)

Therefore, A(1, -1) → A"(2, 0) is the single transformation.

Rule for this transformation which maps triangle C to A will be defined by,

Rotation of triangle A across a point (1.5, -0.5) by 180°.

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2 years ago

Jenna earns $7.75 per hour working at the mall. If Jenna worked hours 22 1/5 last week, how much did she earn before taxes were

drek231 [11]

Jenna would earn gross pay of $172.05

7 0

2 years ago