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2 years ago

If l is parallel to m, find x and y

Mathematics

1 answer:

soldier1979 [14.2K]2 years ago

6 0

Answer:

x = 24, y = 15

Step-by-step explanation:

Find attached the step-by-step solution.

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Solve for x.<br> A. 6/3<br><br> B. 5<br><br> C. 12<br><br> D. 12.5

belka [17]

Answer:

x=12

Step-by-step explanation:

Using Proportion:

$\tt \cfrac{x}{16} =\cfrac{9}{x}$

$\tt x^2=9(16)$

$\tt x=12$

3 0

1 year ago

Solve each if the given equalities<br>2x + 5<-3

Zarrin [17]

Answer:

x < − 4

Step-by-step explanation:

Solve by:

Factoring, substituting for variables, then simplify!

Hope this helps!!

8 0

2 years ago

The null space for the matrix is spanned by the vector:_______

mafiozo [28]

Answer:

True

Step-by-step explanation:

The null space of matrix is set of all solutions to matrix. The linearly independent vectors forms subset which are spanned and forms the null space. The null space of vector can be found by reducing its echelon. The non zero rows formed are the null spaces of matrix.

8 0

2 years ago

when 6 is subtracted from the square of a number, the result is 5 times the number. Find the negative solution.

sveta [45]

When 6 is subtracted from the square of a number, the result is 5 times the number, then the negative solution is -1

<h3><u>Solution:</u></h3>

Given that when 6 is subtracted from the square of a number, the result is 5 times the number

To find: negative solution

Let "a" be the unknown number

Let us analyse the given sentence

square of a number = $a^2$

6 is subtracted from the square of a number = $a^2 - 6$

5 times the number = $5 \times a$

<em><u>So we can frame a equation as:</u></em>

6 is subtracted from the square of a number = 5 times the number

$a^2 - 6 = 5 \times a\\\\a^2 -6 -5a = 0\\\\a^2 -5a -6 = 0$

<em><u>Let us solve the above quadratic equation</u></em>

For a quadratic equation $ax^2 + bx + c = 0$ where $a \neq 0$

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Here in this problem,

$a^2-5 a-6=0 \text { we have } a=1 \text { and } b=-5 \text { and } c=-6$

Substituting the values in above quadratic formula, we get

$\begin{array}{l}{a=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(1)(-6)}}{2 \times 1}} \\\\ {a=\frac{5 \pm \sqrt{25+16}}{2}=\frac{5 \pm \sqrt{49}}{2}} \\\\ {a=\frac{5 \pm 7}{2}}\end{array}$