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ladessa [460]
3 years ago
11

Brainliest if correct

Mathematics
1 answer:
slava [35]3 years ago
5 0

Answer:

-8

2

12

52

Step-by-step explanation:

y = 10x + 2

x = -1, y = 10(-1) + 2 = -8

x - 0, y = 0 + 2 = 2

x = 1, y = 10 + 2 = 12

x = 5, y = 10*5 + 2 = 52

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Please help me on this geometry i’ll mark u the brainliest
kicyunya [14]

Answer:

<h2>               XN = 6</h2>

Step-by-step explanation:

If XY is bisector of angle AXN then:   \dfrac{YN}{AY}=\dfrac{XN}{AX}

\dfrac4{12}=\dfrac{XN}{18}\\\\XN=\dfrac4{12}\cdot18\\\\XN=6                                                            

5 0
3 years ago
Given: measure of angle KJ = 124°, measure of IC =38°<br><br> Find: m∠CQJ, m∠LIJ.
VLD [36.1K]

Answer:

Part 1) m

Part 2) m

Step-by-step explanation:

Part 1) Find the measure of angle CQJ

we know that

The measure of the interior angle is the semi-sum of the arcs comprising it and its opposite.

m

arc\ IK+arc\ CJ=360\°-(arc\ IC+arc\ KJ)

substitute the values

arc\ IK+arc\ CJ=360\°-(38\°+124\°)=198\°

m

Part 2) Find the measure of angle LIJ

step 1

Find the measure of angle IJL

we know that

The inscribed angle is half that of the arc it comprises.

m

substitute the values

m

step 2

Find the measure of angle ILJ

we know that

The measurement of the external angle is the semi-difference of the arcs it encompasses.

m

substitute the values

m

step 3

Find the measure of angle LIJ

Remember that the sum of the interior angles of a triangle must be equal to 180 degrees

In the triangle LIJ

m

substitute the values

m

m

3 0
3 years ago
I have $1000 invested in a coin worth $43000. If the coin reaches $100000 how much will I have invested?
Pavel [41]

Answer:

$2326

Step-by-step explanation:

I have $1000 invested in a coin worth $43000. If the coin reaches $100000 how much will I have invested?

To solve the above question, we have:

$43,000 = $1000

$100,000 = x

Cross Multiply

43000 × x = 100,000 × 1000

x = 100000 × 1000/43000

x = $2325.5813953

Approximately = $2326

Therefore, If the coin reaches $100000 you would have invested $2326

8 0
3 years ago
Daily Distance Larry Biked (Km)
notsponge [240]

Answer:

The most common distance that Larry biked was of 6 Km.

Step-by-step explanation:

The most common distance that Larry biked is the number, in Km, which appears most frequently in this table.

The number which appears the most frequently is:

6 km

This means that the most common distance that Larry biked was of 6 Km.

5 0
3 years ago
Find the area under the standard normal probability distribution between the following pairs of​ z-scores. a. z=0 and z=3.00 e.
prohojiy [21]

Answer:

a. P(0 < z < 3.00) =  0.4987

b. P(0 < z < 1.00) =  0.3414

c. P(0 < z < 2.00) = 0.4773

d. P(0 < z < 0.79) = 0.2852

e. P(-3.00 < z < 0) = 0.4987

f. P(-1.00 < z < 0) = 0.3414

g. P(-1.58 < z < 0) = 0.4429

h. P(-0.79 < z < 0) = 0.2852

Step-by-step explanation:

Find the area under the standard normal probability distribution between the following pairs of​ z-scores.

a. z=0 and z=3.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 3.00) = 0.9987

Thus;

P(0 < z < 3.00) = 0.9987 - 0.5

P(0 < z < 3.00) =  0.4987

b. b. z=0 and z=1.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 1.00) = 0.8414

Thus;

P(0 < z < 1.00) = 0.8414 - 0.5

P(0 < z < 1.00) =  0.3414

c. z=0 and z=2.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 2.00) = 0.9773

Thus;

P(0 < z < 2.00) = 0.9773 - 0.5

P(0 < z < 2.00) = 0.4773

d.  z=0 and z=0.79

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 0.79) = 0.7852

Thus;

P(0 < z < 0.79) = 0.7852- 0.5

P(0 < z < 0.79) = 0.2852

e. z=−3.00 and z=0

From the standard normal distribution tables,

P(Z< -3.00) = 0.0014  and P(Z< 0) = 0.5

Thus;

P(-3.00 < z < 0 ) = 0.5 - 0.0013

P(-3.00 < z < 0) = 0.4987

f. z=−1.00 and z=0

From the standard normal distribution tables,

P(Z< -1.00) = 0.1587  and P(Z< 0) = 0.5

Thus;

P(-1.00 < z < 0 ) = 0.5 -  0.1586

P(-1.00 < z < 0) = 0.3414

g. z=−1.58 and z=0

From the standard normal distribution tables,

P(Z< -1.58) = 0.0571  and P(Z< 0) = 0.5

Thus;

P(-1.58 < z < 0 ) = 0.5 -  0.0571

P(-1.58 < z < 0) = 0.4429

h. z=−0.79 and z=0

From the standard normal distribution tables,

P(Z< -0.79) = 0.2148  and P(Z< 0) = 0.5

Thus;

P(-0.79 < z < 0 ) = 0.5 -  0.2148

P(-0.79 < z < 0) = 0.2852

8 0
3 years ago
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