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3 years ago

If one of two parallel lines is perpendicular to another line, the other of the two lines is ________ to the third line.

Mathematics

2 answers:

eduard3 years ago

7 0

The line is perpendicular to the third line

lidiya [134]3 years ago

6 0

I think the answer is B. perpendicular

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Complete the statement. Trapezoids are parallelograms

Lady bird [3.3K]

Answer:

Step-by-step explanation:

(A) The given statement is:A square is a parallelogram-------A square is always a parallelogram (By properties of parallelogram)(B) The given statement is:A rectangle is a trapezoid----------A rectangle is never a trapezoid.(C) The given statement is:A rhombus is a square--------A rhombus is sometimes a square.(D) The given statement is:A quadrilateral is a kite--------A quadrilateral is sometimes a kite.

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3 years ago

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The base of a triangular prism is an isosceles right triangle with a hypotenuse of 72−−√ centimeters. The height of the prism is

Amiraneli [1.4K]

Answer:

Step-by-step explanation:

Surface area = lateral area + 2(area of base)

Lateral area = perimeter of base * height.

Because it is a isosceles right triangle, both sides are equal.

$x^{2} +x^{2}$ = 72

2 $x^{2}$ = 72. Divide both sides by 2

$x^{2}$ = 36. Square both sides.

x = 6.

So the perimeter of the base = 6 + 6 + $\sqrt{72}$ = 20.485281374239

Lateral area = 20.485281374239 * 7 = 143.397 $cm^{2}$

Area of base is (1/2)base * height.

(1/2)(6)(6) = 18

Using the surface area formula

surface area = 143.397 + 2(18) = 179.4 $cm^{3}$

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3 years ago

A store is having a sale where winter clothes are 60% of the original price. A sweater is on sale for $30 . What was the origina

Flura [38]

$x-price\ before\ sale\\60\%=0.6\\100\%=1\\\\From\ proportion:\\\\0.6\ \ \ \ \ -\ \ \ \ 30\\1\ \ \ \ \ \ \ \ -\ \ \ \ x\\\\0.6x=30\ \ |:0.6\\x=50\\\\Original\ price\ was\ \$50.$

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3 years ago

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Line p has an equation of y=-8x+6. Line q, which is perpendicular to line p, includes the point (2,–2). What is the equation of

Arturiano [62]

Answer:

$y = \frac{1}{8} x - 2 \frac{1}{4}$

Step-by-step explanation:

<u>Slope-intercept </u><u>form</u>

y= mx +c, where m is the slope and c is the y-intercept

Line p: y= -8x +6

slope= -8

The product of the slopes of perpendicular lines is -1. Let the slope of line q be m.

m(-8)= -1

m= -1 ÷(-8)

m= ⅛

Substitute m= ⅛ into the equation:

y= ⅛x +c

To find the value of c, substitute a pair of coordinates that the line passes through into the equation.

When x= 2, y= -2,

-2= ⅛(2) +c

$- 2 = \frac{1}{4} + c$

$c = - 2 - \frac{1}{4}$

$c = - 2 \frac{1}{4}$

Thus, the equation of line q is $y = \frac{1}{8} x - 2 \frac{1}{4}$ .

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2 years ago

A juggler tosses a ball into the air . The balls height, h and time t seconds can be represented by the equation h(t)= -16t^2+40

malfutka [58]

PART A

The given equation is

$h(t) = - 16 {t}^{2} + 40t + 4$

In order to find the maximum height, we write the function in the vertex form.

We factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + 4$

We add and subtract

$- 16(- \frac{5}{4} )^{2}$

to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + - 16( - \frac{5}{4})^{2} - -16( - \frac{5}{4})^{2} + 4$

We again factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4})^{2} ) - -16( - \frac{5}{4})^{2} + 4$

This implies that,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4}) ^{2} ) + 16( \frac{25}{16}) + 4$

The quadratic trinomial above is a perfect square.

$h(t) = - 16 ( t- \frac{5}{4}) ^{2} +25+ 4$

This finally simplifies to,

$h(t) = - 16 ( t- \frac{5}{4}) ^{2} +29$

The vertex of this function is

$V( \frac{5}{4} ,29)$

The y-value of the vertex is the maximum value.

Therefore the maximum value is,

$29$

PART B

When the ball hits the ground,

$h(t) = 0$

This implies that,

$- 16 ( t- \frac{5}{4}) ^{2} +29 = 0$

We add -29 to both sides to get,

$- 16 ( t- \frac{5}{4}) ^{2} = - 29$

This implies that,

$( t- \frac{5}{4}) ^{2} = \frac{29}{16}$

$t- \frac{5}{4} = \pm \sqrt{ \frac{29}{16} }$

$t = \frac{5}{4} \pm \frac{ \sqrt{29} }{4}$

$t = \frac{ 5 + \sqrt{29} }{4} = 2.60$

or

$t = \frac{ 5 - \sqrt{29} }{4} = - 0.10$

Since time cannot be negative, we discard the negative value and pick,

$t = 2.60s$

8 0

3 years ago