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3 years ago

Plz help me I’m confused on to how to do this problem

Mathematics

2 answers:

aliina [53]3 years ago

7 0

67 multiplied by 3 is 201 but 67 and 3 are prime numbers so yeah

arsen [322]3 years ago

3 0

What the problem is asking is to divide 402 into two numbers. Since it already gave one number, 201, what you have to do is divide 402 by 201. 402 divided 201 equals 2. Hope I helped :)

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(no links or files) Solve this equation using the quadratic formula. This question should be solved both ways. (1) Quadratic For

meriva

Quadratic formula is x = -b+ or- sq rt b^2-4ac / 2a
a=2 b=5 c=-3
-5 +or- sqrt 5^2-4(2)(-3) / 2(2)
-5 +or- sqrt 49/ 4
-5 + 7 /4 = 2/4 = 1/2
-5 - 7 /4 = -12/4 = -3

Factoring a*c is 2*-3 =-6
Factors of -6 that add to 5 are 6 and -1
Split 5x into +6x-1x
2x^2+6x-1x-3 and group
2x(x+3)-1(x+3)
(x+3)(2x-1)=0
x+3=0 gives x=-3
2x-1=0 gives x=1/2

4 0

3 years ago

What is the simplified fraction of 13/24

adoni [48]

$\frac{13}{26}$ divide the numerator and denominator by 13

= $\frac{1}{2}$

8 0

3 years ago

Solve the system using substitution x+5y=0 3y+2x=-21

Crazy boy [7]

Answer:

$\left \{ {{x = -15} \atop {y=3}} \right.$

Step-by-step explanation:

$\left \{ {{x + 5y = 0} (1) \atop {3y + 2x = -21}(2)} \right. \left \{ {{x = -5y} \atop {3y + 2x = -21}} \right. \left \{ {{x = -5y} \atop {3y + 2*(-5y) = -21}} \right. \left \{ {{x = -5y} \atop {3y - 10y = -21}} \right. \\ \left \{ {{x = -5y } \atop {-7y = -21}} \right. \left \{ {{x = -5y} \atop {y= 3 }} \right. \left \{ {{x = -5 * 3} \atop {y=3}} \right.=> \left \{ {{x=-15} \atop {y=3}} \right.$

5 0

2 years ago

Read 2 more answers

The simple interest due on a 6-month loan of $2,500 is $125. What is the monthly payment on the loan?

Elina [12.6K]

A person borrows $2500 and they must pay this amount back, plus the $125 in simple interest. In total, they must pay back 2500+125 = 2625 dollars.

Divide this amount over 6 months
2625/6 = 437.50

The monthly payment is $437.50, so the answer is choice B

4 0

3 years ago

Rewrite the system of linear equations as a matrix equation AX = B.

iren2701 [21]

Answer:

$\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]$

Step-by-step explanation:

Let's find the answer.

Because we have 3 equations and 3 variables (x1, x2, x3) a 3x3 matrix (A) can be constructed by using their respectively coefficients.

Equations:

Eq. 1 : x1 + 2x2 + 5x3 = 5

Eq. 2 : x1 + x2 + x3 = 6

E1. 3 : 4x1 + 6x2 + 5x3 = 7

Coefficients for x1 ; x2 ; x3

From eq. 1 : 1 ; 2 ; 5

From eq. 2 : 1 ; 1 ; 1

From eq. 3 : 4 ; 6 ; 5

So matrix A is:

$\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]$

And the vector of vriables (X) is:

$\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]$

Now we can find the resulting vector (B) using the 'resulting values' from each equation:

$\left[\begin{array}{ccc}5\\6\\7\end{array}\right]$

In conclusion, AX=B is:

$\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]$

7 0

3 years ago