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Varvara68 [4.7K]
3 years ago
6

B(a+b)+a; given a =9 and b=4

Mathematics
2 answers:
Harman [31]3 years ago
8 0

b(a+b) + a = 4(9+4) + 9

4(9+4) + 9 = 61

So the answer is 61

Hope this helps

Anna007 [38]3 years ago
4 0

To solve this expression, we simply have to substitute in the given values for a and b into the expression. This means we should replace every a with a 9 and every b with a 4, and then simplify from there. This is modeled below:

b(a+b) + a

4(9+4) + 9

To simplify, we must use the order of operations, which is outlined by PEMDAS. This states that we should simplify what is in parentheses first, then exponents next, after that multiplication and division, and finally addition or subtraction. This lets us know that we should solve what is in the parentheses first in our example.

4(9+4) + 9

4(13) + 9

Because the 4 and 13 are being multiplied together, we should compute that next because it is the next letter in PEMDAS in the expression.

52 + 9

Finally, we can add these two numbers together to find our answer.

61

Therefore, your answer is 61.

Hope this helps!

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Table 1 contains outputs of the function f(x)=b^x for some x values and Tables 2 contains outputs of the function g(x)=logb (x)
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3 years ago
The data below represents the number of customers at each Slurpee Sam's Spaghetti Shop. 24, 25, 29, 30, 31, 31, 32, 34, 34. Whic
NeX [460]

Answer:

A summary of the data is as follows;

1. Maximum = 34

2. Minimum = 24

3. The 1st quartile is 27

4. The median is or the 2nd quartile 34

5. The 3rd quartile is 33

The data is summarized in the attached box plot

Step-by-step explanation:

24, 25, 29, 30, 31, 31, 32, 34, 34

From the given data, we have the five values of a box and whisker plot as follows;

1. Maximum = 34

2. Minimum = 24

3. The 1st quartile is the \frac{9 + 1}{4} th term or 27

4. The median (2nd quartile) is the \frac{9 + 1}{2} th term or 34

5. The 3rd quartile is the \frac{3(9 + 1)}{4} th term or 33

Hence the attached box plot summarizes the data.

6 0
3 years ago
Brandon earns $6 for every shirt he sell.which equation could Brandon use to determine the amount of money he earns for any numb
maw [93]
I got B. Hope it helps 
8 0
3 years ago
Square root of 2tanxcosx-tanx=0
kobusy [5.1K]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/3242555

——————————

Solve the trigonometric equation:

\mathsf{\sqrt{2\,tan\,x\,cos\,x}-tan\,x=0}\\\\ \mathsf{\sqrt{2\cdot \dfrac{sin\,x}{cos\,x}\cdot cos\,x}-tan\,x=0}\\\\\\ \mathsf{\sqrt{2\cdot sin\,x}=tan\,x\qquad\quad(i)}


Restriction for the solution:

\left\{ \begin{array}{l} \mathsf{sin\,x\ge 0}\\\\ \mathsf{tan\,x\ge 0} \end{array} \right.


Square both sides of  (i):

\mathsf{(\sqrt{2\cdot sin\,x})^2=(tan\,x)^2}\\\\ \mathsf{2\cdot sin\,x=tan^2\,x}\\\\ \mathsf{2\cdot sin\,x-tan^2\,x=0}\\\\ \mathsf{\dfrac{2\cdot sin\,x\cdot cos^2\,x}{cos^2\,x}-\dfrac{sin^2\,x}{cos^2\,x}=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left(2\,cos^2\,x-sin\,x \right )=0\qquad\quad but~~cos^2 x=1-sin^2 x}

\mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\cdot (1-sin^2\,x)-sin\,x \right]=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2-2\,sin^2\,x-sin\,x \right]=0}\\\\\\ \mathsf{-\,\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}\\\\\\ \mathsf{sin\,x\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}


Let

\mathsf{sin\,x=t\qquad (0\le t


So the equation becomes

\mathsf{t\cdot (2t^2+t-2)=0\qquad\quad (ii)}\\\\ \begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{2t^2+t-2=0} \end{array}


Solving the quadratic equation:

\mathsf{2t^2+t-2=0}\quad\longrightarrow\quad\left\{ \begin{array}{l} \mathsf{a=2}\\ \mathsf{b=1}\\ \mathsf{c=-2} \end{array} \right.


\mathsf{\Delta=b^2-4ac}\\\\ \mathsf{\Delta=1^2-4\cdot 2\cdot (-2)}\\\\ \mathsf{\Delta=1+16}\\\\ \mathsf{\Delta=17}


\mathsf{t=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{2\cdot 2}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{4}}\\\\\\ \begin{array}{rcl} \mathsf{t=\dfrac{-1+\sqrt{17}}{4}}&\textsf{ or }&\mathsf{t=\dfrac{-1-\sqrt{17}}{4}} \end{array}


You can discard the negative value for  t. So the solution for  (ii)  is

\begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{t=\dfrac{\sqrt{17}-1}{4}} \end{array}


Substitute back for  t = sin x.  Remember the restriction for  x:

\begin{array}{rcl} \mathsf{sin\,x=0}&\textsf{ or }&\mathsf{sin\,x=\dfrac{\sqrt{17}-1}{4}}\\\\ \mathsf{x=0+k\cdot 180^\circ}&\textsf{ or }&\mathsf{x=arcsin\bigg(\dfrac{\sqrt{17}-1}{4}\bigg)+k\cdot 360^\circ}\\\\\\ \mathsf{x=k\cdot 180^\circ}&\textsf{ or }&\mathsf{x=51.33^\circ +k\cdot 360^\circ}\quad\longleftarrow\quad\textsf{solution.} \end{array}

where  k  is an integer.


I hope this helps. =)

3 0
3 years ago
Please help !!
Maurinko [17]
Answer: The number of ceral boxes per day.

Because the slope will be [rise of the grah of the model function] / [run of the graph of the model function].

rise = [increase in the number of cereal boxes sold]

run = [days elapsed ]

rise / run = number of cereal boxes sold per day


8 0
4 years ago
Read 2 more answers
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