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3 years ago

7

Put a check by the three numbers that are prime numbers. 1 7 15 21

2 answers:

Mrac [35]3 years ago

7 0

Seven is the only prime number here

Aleks04 [339]3 years ago

4 0

Answer:

1

7 ✔️

15

21

Only 7 is a prime number in this question.

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A circle in the xy plane has a center at (3/4, 1/2)

Leokris [45]

Answer:

(x - 3/4)^2 + (y - 1/2)^2 = r^2

<em>Brainliest, please!</em>

Step-by-step explanation:

Equation of a circle: (x - h)^2 + (y - k)^2 = r^2

(x - 3/4)^2 + (y - 1/2)^2 = r^2

We do not have enough information to find the radius. Were you given another point that lies on the circle?

3 0

2 years ago

the hight, the feet, of a javelin t seconds afther it is moldeled by the function (t) =4t^2 + 24t + 6. Afther how many seconds d

ruslelena [56]

The max height will be reached after 3 seconds.

You have a typo in your equation. For this type of quadratic equation, it always starts out with a negative coefficient of the t squared term.

To find the x-coordinate of the vertex, we just need to use the expression: -b/2a.

Then, we have:

-24 / (2(-4))

-24 / -8

3

8 0

2 years ago

Surface area of prisms need help ASAP I will give brainliest

trasher [3.6K]

Step-by-step explanation:

A=2πr(r+h)

A=6.28×14mm(+29mm)

A=77.92(14mm+29mm)

A=77.92×43

A=3350.56 mm

Hope it helps

KC

6 0

1 year ago

find the equation of the line passing through the given point and perpendicular to the given equation write your answer in slope

Paraphin [41]

Slope-intercept form:

y = mx + b "m" is the slope, "b" is the y-intercept

For lines to be perpendicular, their slopes have to be the opposite/negative reciprocals (flipped sign and number)

For example:

slope is 2

perpendicular line's slope is -1/2

slope is -2/3

perpendicular line's slope is 3/2

3.) y = 2x - 2

The given line's slope is 2, so the perpendicular line's slope is -1/2

$y=-\frac{1}{2}x+b$ To find "b", plug in the point (-5 , 5) into the equation

$5=-\frac{1}{2}(-5)+b$

$5=\frac{5}{2}+b$ Subtract 5/2 on both sides

$5-\frac{5}{2}=b$ Make the denominators the same

$\frac{10}{2}-\frac{5}{2}=b$

$\frac{5}{2}=b$

$y=-\frac{1}{2}x+\frac{5}{2}$

4.) -6x + 5y = -10 Get "y" by itself, add 6x on both sides

5y = -10 + 6x Divide 5 on both sides

$y=-2+\frac{6}{5}x$

The given line's slope is 6/5, so the perpendicular line's slope is -5/6.

$y=-\frac{5}{6}x+b$ Plug in (-2, 5)

$5 = -\frac{5}{6}(-2)+b$

$5=\frac{10}{6}+b\\ 5=\frac{5}{3}+b$ Subtract 5/3 on both sides

$5-\frac{5}{3} =b$ Make the denominators the same

$\frac{15}{3}-\frac{5}{3}=b\\\frac{10}{3} =b$

$y = -\frac{5}{6}x+\frac{10}{3}$

7.) Perpendicular line's slope is -2

y = -2x + b Plug in (1,4)

4 = -2(1) + b

4 = -2 + b

6 = b

y = -2x + 6

8.) Perpendicular line's slope is -1/4

$y = -\frac{1}{4}x+b$ Plug in (-5 , 2)

$2=-\frac{1}{4}(-5)+b$

$2 = \frac{5}{4}+b$ Subtract 5/4 on both sides

$2-\frac{5}{4}=b$ Make the denominators the same

$\frac{8}{4}-\frac{5}{4}=b$

$\frac{3}{4}=b$

$y=-\frac{1}{4}x+\frac{3}{4}$

5 0

2 years ago

(1-sin^2x) (1+tan^2x)=1

Sphinxa [80]

$\tt (1 - sin ^2x)\times(1+tan^2x) = 1$

L.H.S

$\tt cos^2x \times sec^2x$

$\tt cos^2x \times \frac{1}{cos^2x}$

$\tt \cancel{ cos^2x} \times\cancel{ \frac{1}{cos^2x}}$

$\tt 1$

R.H.S

Hence proved

5 0

2 years ago

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