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tigry1 [53]
3 years ago
8

Line segment t and line segment r are parallel. Both line segments are reflected across the y–axis to form line segments t' and

r'. . What is the distance, in units, between line segmentst' and r'? Select ONE correct answer.
Mathematics
2 answers:
alexgriva [62]3 years ago
7 0

Answer:

i think its 2 i dont know im doing the test rn

Step-by-step explanation:

sorry if its wrong

suter [353]3 years ago
6 0

Answer:

The answer is below

Step-by-step explanation:

The question is not complete, but I would show you how to answer it.

Transformation is the movement of a point from its initial location to a new location. Types of transformation are rotation, translation, reflection and dilation.

Reflection is the flipping of a figure about an axis. Reflection is a rigid transformation, that is it preserves the shape and size of the figure.

If the distance between Line segment t and line segment r is 5 units and then both line segments are reflected across the y–axis to form line segments t' and r'. The distance between the newly formed line segments t' and r' would still be 5 units

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✓ 10
liraira [26]

Answer:

Decrease in dollars is $555.00, $6845.00 was in his account at the end of last year.

Step-by-step explanation:

7400 times 7.5/100

=74 times 7.5

=$555.00

7400-555=$6845.00

7 0
3 years ago
Got half not sure of the rest through​
bezimeni [28]

Answer:

-1/12

Step-by-step explanation:

make the denominators congruent

8/12 and 9/12

then subtract

-1/12

6 0
3 years ago
How do i solve this problem
mamaluj [8]
As they are similar corresponding sides are in the same ratio, so

18/15 = x / 4

x = 4*18 / 15
x = 4.8 answer
7 0
3 years ago
Find the derivative.
Aleksandr [31]

Answer:

Using either method, we obtain:  t^\frac{3}{8}

Step-by-step explanation:

a) By evaluating the integral:

 \frac{d}{dt} \int\limits^t_0 {\sqrt[8]{u^3} } \, du

The integral itself can be evaluated by writing the root and exponent of the variable u as:   \sqrt[8]{u^3} =u^{\frac{3}{8}

Then, an antiderivative of this is: \frac{8}{11} u^\frac{3+8}{8} =\frac{8}{11} u^\frac{11}{8}

which evaluated between the limits of integration gives:

\frac{8}{11} t^\frac{11}{8}-\frac{8}{11} 0^\frac{11}{8}=\frac{8}{11} t^\frac{11}{8}

and now the derivative of this expression with respect to "t" is:

\frac{d}{dt} (\frac{8}{11} t^\frac{11}{8})=\frac{8}{11}\,*\,\frac{11}{8}\,t^\frac{3}{8}=t^\frac{3}{8}

b) by differentiating the integral directly: We use Part 1 of the Fundamental Theorem of Calculus which states:

"If f is continuous on [a,b] then

g(x)=\int\limits^x_a {f(t)} \, dt

is continuous on [a,b], differentiable on (a,b) and  g'(x)=f(x)

Since this this function u^{\frac{3}{8} is continuous starting at zero, and differentiable on values larger than zero, then we can apply the theorem. That means:

\frac{d}{dt} \int\limits^t_0 {u^\frac{3}{8} } } \, du=t^\frac{3}{8}

5 0
3 years ago
Which of the following equations has both 1 and −3 as solutions?
soldi70 [24.7K]
The answer is B)

x2+2x−3=0

Factor left side of equation.

(x−1)(x+3)=0

Set factors equal to 0.

x−1=0 or x+3=0

x=1

or

x=−3

5 0
3 years ago
Read 2 more answers
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