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maria [59]
3 years ago
13

Seventeen minus 6 times 10 divided two plus twelve

Mathematics
2 answers:
Furkat [3]3 years ago
6 0
17-6×10÷2+12
Apply bodmas
=17-6×5+12
=17-30+12
=17-18
=-1
anyanavicka [17]3 years ago
6 0

Answer:

-1

Step-by-step explanation:

Using PEMDAS

17-6*10/2+12

17-30+12

equals -1

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Peggy had three times as many quarters as nickels. She had $1.60 in all. How many nickels and how many quarters did she have? Wh
svlad2 [7]
Let q and n represent the number of quarters and nickels respectively.

q=3n and .05n+.25q=1.6  These are the conditions in mathematical terms.

To solve, use the value of q from the first equation in the second equation to get:

.05n+.25(3n)=1.6  carry out indicated multiplication on left side

.05n+.75n=1.6  combine like terms on left side

.8n=1.6  divide both sides by .8

n=2, since q=3n

q=6

So Peggy had 2 nickels and 6 quarters.
3 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
A circle has a circumference of 907.46.<br> What is the diameter of the circle?
Anna007 [38]

\bf \textit{circumference of a circle}\\\\ C=\pi d~~ \begin{cases} d=diameter\\[-0.5em] \hrulefill\\ C=907.46 \end{cases}\implies 907.46=\pi d\implies \cfrac{907.46}{\pi }=d \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill 288.85\approx d~\hfill

7 0
3 years ago
Some number was divided into 104.13. This quotient was multiplied by 4, after which the resulting product was added to 5. Given
SVEN [57.7K]

Answer:

The initial number is <u>-1218.321</u>.

Step-by-step explanation:

Let the initial number be 'x'.

The number 'x' is divided into 104.13.

So, we divide the number 'x' by 104.13. This gives,

\dfrac{x}{104.13}

Now, the quotient is multiplied by 4. So, this means we need to multiply 4 to the fraction above. This gives,

\dfrac{x}{104.13}\times 4\\\\\frac{4x}{104.13}

Now, 5 is added to the result. This gives,

\frac{4x}{104.13}+5

Now, as per question:

\frac{4x}{104.13}+5=-41.8

Now, solving for 'x', we add -5 both sides. This gives,

\frac{4x}{104.13}+5-5=-41.8-5\\\\\frac{4x}{104.13}=-46.8\\\\4x=-46.8\times 104.13\\\\4x=-4873.284\\\\x=\frac{-4873.284}{4}=-1218.321

Therefore, the initial number is -1218.321.

6 0
3 years ago
If the interest of sum of money is 6 yrs in 3/8 of the principal . what is the rate of interest?​
REY [17]

Answer :

R = 6.25%

Solution,

Let P is the principal. It means interest will be :

I=3/8 P

The formula for the interest is given by :

I=\dfrac{PRT}{100}\\\\\dfrac{3}{8}P=\dfrac{PR\times 6}{100}\\\\\dfrac{3}{8}=\dfrac{6R}{100}\\\\R=\dfrac{300}{8\times 6}\\\\=6.25\%

Hence, the rate of interest is 6.25%.

6 0
3 years ago
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