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2 years ago

Babatunde draws a drinking glass on graph paper using the scale shown below. The glass has a length of 888 units in the drawing.

Mathematics

2 answers:

Helen [10]2 years ago

8 0

Answer: Is there a picture? It would really help me understand you question if you added a picture.

Step-by-step explanation:

Harlamova29_29 [7]2 years ago

4 0

10 cm

The glass is 8 units. 2 units = 2.5 cm. Multiply 8 by 2.5 and divide it by 2, and you get 10 cm

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HELP

ladessa [460]

Use the Pythagorean Theorem:
A²+B²=C²
6²+5²=C²
36+25=C²
61=C²
√61=C
So the answer is C.

7 0

2 years ago

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This table shows values that represent an exponential function.

Sati [7]

Answer:

y=21

Step-by-step explanation:

3 0

2 years ago

Please answer correctly !!!!!! Will mark brainliest !!!!!!!!!!!!!

Leni [432]

Answer:

At the time of launch height of the object was 60 meters.

Step-by-step explanation:

An object was launched from a platform and its height was modeled by the function,

h(x) = -5x² + 20x + 60

Where x = time or duration after the launch

At the time of launch, x = 0

So, by putting x = 0 in this equation,

h(0) = -5×(0) + 20×(0) + 60

h(0) = 60

Therefore, at the time of launch height of the object was 60 meters.

4 0

3 years ago

A deck of cards contains RED cards numbered 1,2,3 and BLUE cards numbered 1,2,3,4,5,6. Let R be the event of drawing a red card,

Tatiana [17]

Answer:

e. R′

Step-by-step explanation:

Let the Sample Space be a universal set consisting of 3 red and 6 blue cards. Then the event of of getting 4 blue cards will be given

R complement = R` = Universal Set minus Red cards will give blue cards.

a. R OR O

It cannot be this option because we need 4 blue card

b. B AND O

It cannot be this choice as well because 4 is not odd.

c. R OR E

4 is even but we need blue

d. R AND O

Red and odd is again not required

e. R′ = It will give our required result.

f. E′= 4 is even if it is complemented it cannot be obtained and will be left out.

5 0

3 years ago

Match the parabolas represented by the equations with their vertices. y = x2 + 6x + 8 y = 2x2 + 16x + 28 y = -x2 + 5x + 14 y = -

GaryK [48]

Consider all parabolas:

$y = x^2 + 6x + 8,\\y=x^2+6x+9-9+8,\\y=(x^2+6x+9)-1,\\y=(x+3)^2-1.$

When x=-3, y=-1, then the point (-3,-1) is vertex of this first parabola.

$y = 2x^2 + 16x + 28=2(x^2+8x+14),\\y=2(x^2+8x+16-16+14),\\y=2((x^2+8x+16)-16+14),\\y=2((x+4)^2-2)=2(x+4)^2-4.$

When x=-4, y=-4, then the point (-4,-4) is vertex of this second parabola.

$y =-x^2 + 5x + 14=-(x^2-5x-14),\\y=-(x^2-5x+\dfrac{25}{4}-\dfrac{25}{4}-14),\\y=-((x^2-5x+\dfrac{25}{4})-\dfrac{25}{4}-14),\\y=-((x-\dfrac{5}{2})^2-\dfrac{81}{4})=-(x-\dfrac{5}{2})^2+\dfrac{81}{4}.$

When x=2.5, y=20.25, then the point (2.5,20.25) is vertex of this third parabola.

$y =-x^2 + 7x + 7=-(x^2-7x-7),\\y=-(x^2-7x+\dfrac{49}{4}-\dfrac{49}{4}-7),\\y=-((x^2-7x+\dfrac{49}{4})-\dfrac{49}{4}-7),\\y=-((x-\dfrac{7}{2})^2-\dfrac{77}{4})=-(x-\dfrac{7}{2})^2+\dfrac{77}{4}.$

When x=3.5, y=19.25, then the point (3.5,19.25) is vertex of this fourth parabola.

$y =2x^2 + 7x +5=2(x^2+\dfrac{7}{2}x+\dfrac{5}{2}),\\y=2(x^2+\dfrac{7}{2}x+\dfrac{49}{16}-\dfrac{49}{16}+\dfrac{5}{2}),\\y=2((x^2+\dfrac{7}{2}x+\dfrac{49}{16})-\dfrac{49}{16}+\dfrac{5}{2}),\\y=2((x+\dfrac{7}{4})^2-\dfrac{9}{16})=2(x+\dfrac{7}{4})^2-\dfrac{9}{8}.$

When x=-1.75, y=-1.125, then the point (-1.75,-1.125) is vertex of this fifth parabola.

$y =-2x^2 + 8x +5=-2(x^2-4x-\dfrac{5}{2}),\\y=-2(x^2-4x+4-4-\dfrac{5}{2}),\\y=-2((x^2-4x+4)-4-\dfrac{5}{2}),\\y=-2((x-2)^2-\dfrac{13}{2})=-2(x-2)^2+13.$

When x=2, y=13, then the point (2,13) is vertex of this sixth parabola.

3 0

3 years ago