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3 years ago

Solve for 2/5 + 3/8= With two possible answeres

1 answer:

6 0

You can change the denominators so they are alike. $\frac{16}{40} + \frac{15}{40} = \frac{31}{40}$
Not sure how there can be two possible answers unless you're mentioning the decimal form:
.775
Tell me if this helps!!

You might be interested in

Express in the form of P/Q i) 0.675 (ii) 0.32

quester [9]

Express in the form of P/Q

i) 0.675

= 675/1000

(ii) 0.32

= 32/100

5 0

3 years ago

School 2x still sucks

garik1379 [7]

Answer: 1

Explanation:

$1\frac{3}{5} = \frac{8}{5}$

$\frac{8}{5} * \frac{5}{8} = \frac{40}{40}$

5 0

3 years ago

Read 2 more answers

You are baking brownies for your class. There are 28 total students in your class and you have baked 23 brownies. Write and solv

barxatty [35]

Answer:

2(28)=23+x

Step-by-step explanation:

There are 28 students and each student needs two brownies. You have to multiply 28 by 2 so you will have to total number on one side.

On the other you need to have the number of brownies you have at the moment and x for the number you need to bake. You want to add them because you will need to put those numbers together to get the number on the opposite side.

28x2=56 so 56=23+x. Subtract the 23 from both sides to get 33=x. x=33

To check plug 33 into x to get 2(28)=23+33. 56=23+33. 56=56

you need 33 more brownies.

So the equation would be 2(28)=23+x.

Hope that helps.

3 0

3 years ago

Thats it, ima be also posting more if i dont get the question

soldi70 [24.7K]

The equation of this straight line is y = 50x + 100.

<h3>What is a linear function?</h3>

A linear function can be defined as a type of function whose equation is graphically represented by a straight line on the cartesian coordinate.

<h3>How to determine the equation?</h3>

In order to determine the equation from this graph of a linear equation, we would find the slope of the straight line by using this formula as follows:

$Slope,\;m = \frac{Change\;in\;y\;axis}{Change\;in\;x\;axis}\\\\Slope,\;m = \frac{y_2\;-\;y_1}{x_2\;-\;x_1}$

Slope, m = 100/2

Slope, m = 50.

Mathematically, the standard form of the equation of a straight line is given by;

y = mx + c

Where:

x and y are the points.
m is the slope.
c is the intercept.

At point (4, 300), we have:

y = mx + c

300 = 50(4) + c

300 = 200 + c

c = 300 - 200

c = 100.

Therefore, the equation of this straight line is y = 50x + 100.

Read more on slope here: brainly.com/question/28312822

#SPJ1

5 0

1 year ago

Consider the matrix A =(1 1 1 3 4 3 3 3 4) Find the determinant |A| and the inverse matrix A^-1.

solong [7]

Answer:

$A)\,\,det(A)=1$

$B)\,\,A^{-1}=\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right]$

Step-by-step explanation:

$det(A) = \left\Bigg|\begin{array}{ccc}1&1&1\\3&4&3\\3&3&4\end{array}\right\Bigg|$

Expanding with first row

$det(A) = \left\Bigg|\begin{array}{ccc}1&1&1\\3&4&3\\3&3&4\end{array}\right\Bigg|\\\\\\det(A)= (1)\left\Big|\begin{array}{cc}4&3\\3&4\end{array}\right\Big|-(1)\left\Big|\begin{array}{cc}3&3\\3&4\end{array}\right\Big|+(1)\left\Big|\begin{array}{cc}3&4\\3&3\end{array}\right\Big|\\\\det(A)=1[16-9]-1[12-9]+1[9-12]\\\\det(A)=7-3-3\\\\det(A)=1$

To find inverse we first find cofactor matrix

$C_{1,1}=(-1)^{1+1}\left\Big|\begin{array}{cc}4&3\\3&4\end{array}\right\Big|=7\\\\C_{1,2}=(-1)^{1+2}\left\Big|\begin{array}{cc}3&3\\3&4\end{array}\right\Big|=-3\\\\C_{1,3}=(-1)^{1+3}\left\Big|\begin{array}{cc}3&4\\3&3\end{array}\right\Big|=-3\\\\C_{2,1}=(-1)^{2+1}\left\Big|\begin{array}{cc}1&1\\3&4\end{array}\right\Big|=-1\\\\C_{2,2}=(-1)^{2+2}\left\Big|\begin{array}{cc}1&1\\3&4\end{array}\right\Big|=1\\\\C_{2,3}=(-1)^{2+3}\left\Big|\begin{array}{cc}1&1\\3&3\end{array}\right\Big|=0\\\\$

$C_{3,1}=(-1)^{3+1}\left\Big|\begin{array}{cc}1&1\\4&3\end{array}\right\Big|=-1\\\\C_{3,2}=(-1)^{3+2}\left\Big|\begin{array}{cc}1&1\\3&3\end{array}\right\Big|=0\\\\\\C_{3,3}=(-1)^{3+3}\left\Big|\begin{array}{cc}1&1\\3&4\end{array}\right\Big|=1\\\\$

Cofactor matrix is

$C=\left[\begin{array}{ccc}7&-3&3\\-1&1&0\\-1&0&1\end{array}\right] \\\\Adj(A)=C^{T}\\\\Adj(A)=\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right] \\\\\\A^{-1}=\frac{adj(A)}{det(A)}\\\\A^{-1}=\frac{\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right] }{1}\\\\A^{-1}=\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right]$

4 0

3 years ago