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2 years ago

What are the coordinates of A after reflecting across the line x = 3?

Mathematics

1 answer:

Pachacha [2.7K]2 years ago

3 0

Answer:

It would be (3,2) since it is being reflected over the line it is already on

Step-by-step explanation:

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Jeffery has grades of 93 and 81 on the first two testes of the quarter. Progress reports are sent home after the third test. He

navik [9.2K]

Answer:

96 or 97

Step-by-step explanation:

5 0

3 years ago

In words, the scalar product of two vectors can be thought of as the product of the magnitude of ~a with the magnitude of the pr

timama [110]

Answer:

67.73625

Step-by-step explanation:

The dot (scalar) product is also the sum of products of corresponding vector components.

~a • ~b = 9·2.97 +6.75·6.075 +0·0 = 27.73 +41.00625 = 67.73625

6 0

3 years ago

What is the volume of this prism if there are no gaps or overlaps in the unit cubes?

777dan777 [17]

Nothing there.

Use this as a tip:

To find the volume of a rectangular prism, multiply its 3 dimensions: length x width x height. The volume is expressed in cubic units.

7 0

2 years ago

What type of triangle is shown in the figure?

Ivan

Exacty two sides are congruent so its an Isosceles triangle

5 0

3 years ago

PLEAZZZZZZZZZZ HELP ASAP

stich3 [128]

QUESTION 1

GIVEN:

$long \: base = 3 \times (short \: base)$

$height = (short \: base) - 2$

$area = 30 {in}^{2}$

IMPORTANT EQUATIONS

$trapezoid \: area = \frac{sum \: of \: bases}{2} \times height$

SOLVE:

$30 = \frac{short + long}{2} (short - 2)$
$30 = \frac{(short + 3short)}{2} \times (short - 2)$

$30 = \frac{4short}{2} (short - 2)$

$30 = 2{(short)}^{2} - 4(short)$

${short}^{2} - 2(short) - 15 = 0$

factor:

$(short - 5)(short + 3) = 0$

$short = 3 \: and \: - 5$
Negative five is not a reasonable answer, so we focus on the positive three and say that is the length of the short base.

so, the answers:

$short \: base = 3 \: in$

$long \: base = 3 \times 3 = 9 \: in$

$height = 3 - 2 = 1 \: in$

QUESTION 2

GIVEN
convert 3ft to inches.

$l = 32 + w$

$area = lw = 130 {in}^{2}$
or

$w = \frac{130}{l}$

substitute

$l = 32 + \frac{130}{l}$

solve for l:

${l}^{2} - 32 l- 130 = 0$
QUESTION 3

$x(x + 2) = 80$

solve for X

${x}^{2} + 2x - 80 = 0$
$(x - 10)(x + 8) = 0$

$x = 10 \: and \: - 8$

take the positive value since it's the only one that makes sense in this context.

so the two values are:

$10 \: and \: 8$

8 0

3 years ago