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zysi [14]
3 years ago
5

What is the midpoint of (0,0.55) and (-3,6)

Mathematics
1 answer:
Contact [7]3 years ago
8 0

Answer:

(-1.5, 3.275)

Step-by-step explanation:

midpoint = ( x1+X2/2 , y1+y2/2)

=(0+(-3)/2, 0.55+6/2)

=( -3/2, 6.55/2)

=( -1.5, 3.275)

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Is 49 a perfect square yes or no
Romashka [77]

Answer:

Yes

Step-by-step explanation:

7 x 7 = 49

4 0
3 years ago
Read 2 more answers
How to solve (x+3)^2+7=29
tangare [24]
X^2 + 9 + 7 = 20
X^2 +16 = 20
X^2 = 20-16
X^2 = 4
Root 4 = 2
X=2
6 0
3 years ago
You need 2b cups of flour for making b loaves of bread. You have 8 cups of flour . Do you have enough for 5 loafs of bread?Expla
lubasha [3.4K]

Answer:

No.  I don't have enough Flour to make 5 loaves of bread.

From the Amount of flour i had i can make 4 loaves only.

Step-by-step explanation:

Given:

Flour required for making loaves of bread = 2b cups

Amount of flour i had = 8 cups

Number of loaves of bread = b

We need to find the we have enough flour to make 5 loafs of bread.

Solution:

We know that;

Amount of flour I had is equal to Flour required for making loaves of bread

framing in equation form we get;

2b=8

Dividing both side by 2 we get;

\frac{2b}{2} =\frac{8}{2}

b=4

Hence From the Amount of flour i had i can make 4 loaves only.

Hence I don't have enough Flour to make 5 loaves of bread.

3 0
3 years ago
A. Solve for y.<br> B. What is the measure of the missing angles?
Alik [6]
Answer: A.) y = 15 B.) (5y + 3)° = 78° (4y + 8)° = 68° and 34°

Steps:
180° - 146° = 34°

180 = 34 + (5y + 3) + (4y + 8)
180 - 34 = (5y + 3) + (4y + 8)
146 = (5y + 3) + (4y + 8)
146 = 5y + 3 + 4y + 8
146 = 9y + 11
146 - 11 = 9y
135 = 9y
135/ 9 = y
15 = y

(5y + 3)
5(15) + 3
75 + 3
78
(5y + 3) = 78

(4y + 8)
4(15) + 8
60 + 8
68
68 = (4y + 8)

Check:
68 + 78 + 34 = 180
180 = 180 ✅
5 0
3 years ago
The graph of an exponential function is given. Which of the following is the correct equation of the function?
katen-ka-za [31]

Answer:

If one of the data points has the form  

(

0

,

a

)

, then a is the initial value. Using a, substitute the second point into the equation  

f

(

x

)

=

a

(

b

)

x

, and solve for b.

If neither of the data points have the form  

(

0

,

a

)

, substitute both points into two equations with the form  

f

(

x

)

=

a

(

b

)

x

. Solve the resulting system of two equations in two unknowns to find a and b.

Using the a and b found in the steps above, write the exponential function in the form  

f

(

x

)

=

a

(

b

)

x

.

EXAMPLE 3: WRITING AN EXPONENTIAL MODEL WHEN THE INITIAL VALUE IS KNOWN

In 2006, 80 deer were introduced into a wildlife refuge. By 2012, the population had grown to 180 deer. The population was growing exponentially. Write an algebraic function N(t) representing the population N of deer over time t.

SOLUTION

We let our independent variable t be the number of years after 2006. Thus, the information given in the problem can be written as input-output pairs: (0, 80) and (6, 180). Notice that by choosing our input variable to be measured as years after 2006, we have given ourselves the initial value for the function, a = 80. We can now substitute the second point into the equation  

N

(

t

)

=

80

b

t

to find b:

⎧

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎨

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎩

N

(

t

)

=

80

b

t

180

=

80

b

6

Substitute using point  

(

6

,

180

)

.

9

4

=

b

6

Divide and write in lowest terms

.

b

=

(

9

4

)

1

6

Isolate  

b

using properties of exponents

.

b

≈

1.1447

Round to 4 decimal places

.

NOTE: Unless otherwise stated, do not round any intermediate calculations. Then round the final answer to four places for the remainder of this section.

The exponential model for the population of deer is  

N

(

t

)

=

80

(

1.1447

)

t

. (Note that this exponential function models short-term growth. As the inputs gets large, the output will get increasingly larger, so much so that the model may not be useful in the long term.)

We can graph our model to observe the population growth of deer in the refuge over time. Notice that the graph below passes through the initial points given in the problem,  

(

0

,

8

0

)

and  

(

6

,

18

0

)

. We can also see that the domain for the function is  

[

0

,

∞

)

, and the range for the function is  

[

80

,

∞

)

.

Graph of the exponential function, N(t) = 80(1.1447)^t, with labeled points at (0, 80) and (6, 180).If one of the data points has the form  

(

0

,

a

)

, then a is the initial value. Using a, substitute the second point into the equation  

f

(

x

)

=

a

(

b

)

x

, and solve for b.

If neither of the data points have the form  

(

0

,

a

)

, substitute both points into two equations with the form  

f

(

x

)

=

a

(

b

)

x

. Solve the resulting system of two equations in two unknowns to find a and b.

Using the a and b found in the steps above, write the exponential function in the form  

f

(

x

)

=

a

(

b

)

x

.

EXAMPLE 3: WRITING AN EXPONENTIAL MODEL WHEN THE INITIAL VALUE IS KNOWN

In 2006, 80 deer were introduced into a wildlife refuge. By 2012, the population had grown to 180 deer. The population was growing exponentially. Write an algebraic function N(t) representing the population N of deer over time t.

SOLUTION

We let our independent variable t be the number of years after 2006. Thus, the information given in the problem can be written as input-output pairs: (0, 80) and (6, 180). Notice that by choosing our input variable to be measured as years after 2006, we have given ourselves the initial value for the function, a = 80. We can now substitute the second point into the equation  

N

(

t

)

=

80

b

t

to find b:

⎧

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎨

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎩

N

(

t

)

=

80

b

t

180

=

80

b

6

Substitute using point  

(

6

,

180

)

.

9

4

=

b

6

Divide and write in lowest terms

.

b

=

(

9

4

)

1

6

Isolate  

b

using properties of exponents

.

b

≈

1.1447

Round to 4 decimal places

.

NOTE: Unless otherwise stated, do not round any intermediate calculations. Then round the final answer to four places for the remainder of this section.

The exponential model for the population of deer is  

N

(

t

)

=

80

(

1.1447

)

t

. (Note that this exponential function models short-term growth. As the inputs gets large, the output will get increasingly larger, so much so that the model may not be useful in the long term.)

We can graph our model to observe the population growth of deer in the refuge over time. Notice that the graph below passes through the initial points given in the problem,  

(

0

,

8

0

)

and  

(

6

,

18

0

)

. We can also see that the domain for the function is  

[

0

,

∞

)

, and the range for the function is  

[

80

,

∞

)

.

Graph of the exponential function, N(t) = 80(1.1447)^t, with labeled points at (0, 80) and (6, 180).

Step-by-step explanation:

4 0
3 years ago
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