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3 years ago

10TH GRADE GEOMETRY PLEASE HELP ME!

Mathematics

2 answers:

fredd [130]3 years ago

7 0

The Figures that are shown in the diagram are
Ray CD
and
Segment CD

anygoal [31]3 years ago

5 0

Answer:

B: point D

C: ray CD

E: segment CD

Options 2,3, and 5

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The points (1,1) and (7,3) are vertices of a parallelogram.

mixer [17]

Step-by-step explanation:

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8 0

2 years ago

PLEASE HELP!!!!!!!!! WILL MARK BRAINLIEST!!

svet-max [94.6K]

Answer:

b=-0.0019a + 215

at 2500 ft, b=210.25 degrees F

Step-by-step explanation:

First find the slope. a=altitude, b = temperature

slope is b2-b1 / a2-a1

points are (8200, 199.42) and (4700, 206.07)

(206.07 - 199.42) / (4700-8200)

= 6.65/ -3500

= - 0.0019

we have b = -0.0019a + intercept

plug in a point to find intercept

199.42 = -0.0019*8200 + intercept

intercept = 199.42 + 15.58 = 215

b=-0.0019a + 215

at a=2500 ft, substitute into equation

b= -0.0019*2500 + 215

= -4.75 + 215

= 210.25 degrees F

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4 0

2 years ago

Find the value of x, if angle x measures 102 degrees

Over [174]

The answer is B
Have a good day!

3 0

3 years ago

Read 2 more answers

With this diagram, what could be the values of c and d?

Sedaia [141]

Answer:

idk maybe

Step-by-step explanation:

7 0

3 years ago

Given the following three points, find by the hand the quadratic function they represent (0,6, (2,16, (3,33)

Lisa [10]

Answer:

$f(x) = 4x^2 - 3x + 6$

Step-by-step explanation:

Quadratic function is given as $f(x) = ax^2 + bx + c$

Let's find a, b and c:

Substituting (0, 6):

$6 = a(0)^2 + b(0) + c$

$6 = 0 + 0 + c$

$c = 6$

Now that we know the value of c, let's derive 2 system of equations we would use to solve for a and b simultaneously as follows.

Substituting (2, 16), and c = 6

$f(x) = ax^2 + bx + c$

$16 = a(2)^2 + b(2) + 6$

$16 = 4a + 2b + 6$

$16 - 6 = 4a + 2b + 6 - 6$

$10 = 4a + 2b$

$10 = 2(2a + b)$

$\frac{10}{2} = \frac{2(2a + b)}{2}$

$5 = 2a + b$

$2a + b = 5$ => (Equation 1)

Substituting (3, 33), and c = 6

$f(x) = ax^2 + bx + x$

$33 = a(3)^2 + b(3) + 6$

$33 = 9a + 3b + 6$

$33 - 6 = 9a + 3b + 6 - 6$

$27 = 9a + 3b$

$27 = 3(3a + b)$

$\frac{27}{3} = \frac{3(3a + b)}{3}$

$9 = 3a + b$

$3a + b = 9$ => (Equation 2)

Subtract equation 1 from equation 2 to solve simultaneously for a and b.

$3a + b = 9$

$2a + b = 5$

$a = 4$

Replace a with 4 in equation 2.

$2a + b = 5$

$2(4) + b = 5$

$8 + b = 5$

$8 + b - 8 = 5 - 8$

$b = -3$

The quadratic function that represents the given 3 points would be as follows:

$f(x) = ax^2 + bx + c$

$f(x) = (4)x^2 + (-3)x + 6$

$f(x) = 4x^2 - 3x + 6$

6 0

3 years ago