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3 years ago

Evaluate: 2a + 7b for a = 5 and b = 3

Mathematics

2 answers:

Schach [20]3 years ago

7 0

Answer:

Step-by-step explanation:

Substitute the values for the variables.

$2(5)+7(3)\\10+21\\31$

jonny [76]3 years ago

3 0

hello,

substitute in the numbers

2(5) + 7(3)

10 + 21

= 31

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Please help!! circles combos are too hard

malfutka [58]

Answer:

56 + 53pi

Step-by-step explanation:

Area of small circles:

diameter of small circle: 4cm

forumla to find area of circle: A = pir^2

r is radius = half of diameter -> d/2 = 4 / 2 = 2cm

A = pi (2cm)^2

A = pi (4cm)

A = 4pi

Area of large circle:

diameter of small circle: 4cm

forumla to find area of circle: A = pir^2

r is radius = half of diameter -> d/2 = 14 / 2 = 7cm

A = pi (7cm)^2

A = pi (49cm)

A = 49pi

Area of rectangle:

Area = width x length

Area = 14cm x 4cm

Area = 56cm

Add all three areas:

Area of rectangle + large circle + small circle

56cm + 49pi + 4pi = 56cm + 53pi

7 0

2 years ago

I need help with this question?

Delicious77 [7]

If I remember right the answer is c

5 0

4 years ago

Asap help.! Circle measurements

Margaret [11]

Answer:

Step-by-step explanation:

Here, we want to get the value of x;

To get this, we simply use one of the circle angle and arc properties

Mathematically;

arc MN = 2 * arc MLN

3x + 22 = 134/2

That will be;

3x + 22 = 67

3x = 67-22

3x = 45

x = 45/3

x = 15

5 0

3 years ago

Find two unit vectos that are orthogonal to both [0,1,2] and [1,-2,3]

alekssr [168]

Answer:

Let the vectors be

a = [0, 1, 2] and

b = [1, -2, 3]

( 1 ) The cross product of a and b (a x b) is the vector that is perpendicular (orthogonal) to a and b.

Let the cross product be another vector c.

To find the cross product (c) of a and b, we have

$\left[\begin{array}{ccc}i&j&k\\0&1&2\\1&-2&3\end{array}\right]$

c = i(3 + 4) - j(0 - 2) + k(0 - 1)

c = 7i + 2j - k

c = [7, 2, -1]

( 2 ) Convert the orthogonal vector (c) to a unit vector using the formula:

c / | c |

Where | c | = √ (7)² + (2)² + (-1)² = 3√6

Therefore, the unit vector is

$\frac{[7,2,-1]}{3\sqrt{6} }$

[ $\frac{7}{3\sqrt{6} }$ , $\frac{2}{3\sqrt{6} }$ , $\frac{-1}{3\sqrt{6} }$ ]

The other unit vector which is also orthogonal to a and b is calculated by multiplying the first unit vector by -1. The result is as follows:

[ $\frac{-7}{3\sqrt{6} }$ , $\frac{-2}{3\sqrt{6} }$ , $\frac{1}{3\sqrt{6} }$ ]

In conclusion, the two unit vectors are;

[ $\frac{7}{3\sqrt{6} }$ , $\frac{2}{3\sqrt{6} }$ , $\frac{-1}{3\sqrt{6} }$ ]

and

[ $\frac{-7}{3\sqrt{6} }$ , $\frac{-2}{3\sqrt{6} }$ , $\frac{1}{3\sqrt{6} }$ ]

Hope this helps!

7 0

3 years ago

a) If the null hypothesis is true, you'll get a high P-value. b) If the null hypothesis is true, a P-value of 0.01 will oc

maxonik [38]

Answer:

a) If the null hypothesis is true, you'll get a high P-value. (it depends)

b) If the null hypothesis is true, a P-value of 0.01 will occur about 1% of the time. (false)

c) A P-value of 0.90 means that the null hypothesis has a good chance of being true. (not only has a good chance it has strong evidence)

d) A P-value of 0.90 is strong evidence that the null hypothesis is true.(true)

Step-by-step explanation:

Before i answer this question, you need to understand that p-values give you the clues to identify when you can accept the null hypothesis ( null hypothesis is true) and when you can reject the null hypothesis (null hypothesis is not true).

1. When you get a small p-value (typically ≤ 0.05) values that are less or equal to 0.05, for example 0.01, you reject the null hypothesis (null hypothesis is not true)

2. when you get a large p-value (> 0.05) values that are greater than 0.05, for example 0.94, 0.90. you can accept the null hypothesis because indicates weak evidence against the null hypothesis (null hypothesis is true).

This is the explanation:

a) if the null hypothesis is true you`ll get a high p-value only if the p-value is ≥ 0.05

b) if p value is less or equal to 0.05. Null hypothesis is not true.

c) A P-value of 0.90 means that the null hypothesis has a good chance of being true . It not only has a good chance it is strong evidence that null hypothesis is true.

d) A p-value of 0.90 is strong evidence that null hypothesis is true. p-values that are greater than 0.05 you can accept the null hypothesis (null hypothesis is true).

7 0

3 years ago