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Viefleur [7K]
2 years ago
12

Calculate the area of a triangle with a base of 6 cm and a height of 12 cm

Mathematics
2 answers:
Alex_Xolod [135]2 years ago
7 0

Answer:

36

Step-by-step explanation:

12 x 6 = 72 / 2 = 36 because the area of a triangle is base times height divided by 2

love history [14]2 years ago
6 0
Answer: 36 cm^2

Step-by-Step Explanation:

Base (b) = 6 cm
Height (h) = 12 cm

Area = 1/2 * b * h

Therefore,
= 1/2 * b * h
= 1/2 * 6 * 12
= 1 * 3 * 12
= 3 * 12
= 36

Area = 36 cm^2
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What is the rate of change of y with respect to x for this function?​
lilavasa [31]

Answer:

Given the function: y=f(x) = 3x+2

when x=-2 at the beginning of the interval [-2, 5],

then;

y = 3x+2 begins at

y= 3(-2)+2 = -6+2= -4.

and

when x=5 at the end of the interval [-2, 5],

y = 3x+2 ends up at

y= 3(5)+2 = 15+2= 17.

So,

y has changed -4 to 17, which is a change of 17-(-4)= 17+4 = 21

and x has changed from -2 to 5, which is a change of 5-(-2)=5+2=7

So, the average rate of change of y with respect to x over the interval

[-2, 5] is given by ;

=

Therefore, the average rate of change y with respect to x over the interval is, 3

Step-by-step explanation:

4 0
3 years ago
What is the y- intercept of the line shown?
timofeeve [1]
Y-intercept when x = 0
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3 years ago
Read each of the following situations and select ALL of the ones considered plagiarism.
mrs_skeptik [129]

Answer:

Ashley uses internal citation to give credit for someone else's idea

7 0
3 years ago
△klm, lm=20 sqrt 3 m∠k=105°, m∠m=30° find: kl and km
Anarel [89]

Answer:

KL =  \frac{20\sqrt{6}}{1+\sqrt{3}} = 17.93

MK =  \frac{40\sqrt{3}}{1+\sqrt{3}} = 25.36


Explanation:

According to the Law of Sines:

\frac{a}{sinA}=\frac{b}{sinB}= \frac{c}{sinC}

where:

A, B, and C are angles

a, b, and c are the sides opposite to the angles


First of all, let's find m∠L: the sum of the angles of a triangle is 180°, therefore

m∠K + m∠L + m∠M = 180°

m∠L = 180° - m∠K - m∠M

m∠L = 180° - 105° - 30°

m∠L = 45°


Now, we can apply the Law of Sines to our case (see picture attached):

\frac{LM}{sinK}=\frac{MK}{sinL}=\frac{KL}{sinM}


Let's solve one side at the time:

\frac{LM}{sinK}=\frac{MK}{sinL}

\frac{20\sqrt{3}}{sin(105)}=\frac{MK}{sin(45)}

MK = \frac{20\sqrt{3} }{sin(105)} \cdot sin(45)

MK = \frac{40\sqrt{3} }{1+\sqrt{3} } = 25.36


Similarily:

\frac{LM}{sinK}=\frac{KL}{sinM}

\frac{20\sqrt{3}}{sin(105)}=\frac{KL}{sin(30)}

KL = \frac{20\sqrt{3} }{sin(105)} \cdot sin(30)

KL = \frac{20\sqrt{6}}{1+\sqrt{3}} = 17.93

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3 years ago
Help please find the circumference of the circle
GREYUIT [131]

Answer:

10 i believe

Step-by-step explanation:

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