What must be true if = 30 Ze? Assume a+0,b=0, and y+0.

Mathematics

1 answer:

Aliun [14]3 years ago

5 0

Answer:

Step-by-step explanation:

we can find x/y by multiply 3a/b * 3a/b

x/y = 9a^2/b^2

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For each order pair (x,y) , determine whether it is a solution to the inequality 4x-6y_< - 18

anyanavicka [17]

Solution:

Given:

$4x-6y\leq-18$

To determine if an ordered pair is a solution, we substitute the x and y-values into the inequality and check if it is true.

Thus,

$\begin{gathered} For\text{ the point }(-9,-2) \\ x=-9,y=-2 \\ 4x-6y\leq-18 \\ 4(-9)-6(-2)=-36+12=-24 \\ Since\text{ }-24$

$\begin{gathered} For\text{ the point }(0,3) \\ x=0,y=3 \\ 4x-6y\leq-18 \\ 4(0)-6(3)=0-18=-18 \\ Since\text{ }-18=-18,\text{ then }(0,3)\text{ is a solution} \end{gathered}$

$\begin{gathered} For\text{ the point }(5,-3) \\ x=5,y=-3 \\ 4x-6y\leq-18 \\ 4(5)-6(-3)=20+18=38 \\ Since\text{ }38>-18,\text{ then }(5,-3)\text{ is not a solution} \end{gathered}$

$\begin{gathered} For\text{ the point }(8,5) \\ x=8,y=5 \\ 4x-6y\leq-18 \\ 4(8)-6(5)=32-30=2 \\ Since\text{ }2>-18,\text{ then }(8,5)\text{ is not a solution} \end{gathered}$

Therefore, the correct answers are selected as shown below;

7 0

1 year ago

Can someone help me

Lemur [1.5K]

Multiplying fractions is just multiplying the numerators together and the denominators together.

$\sf\dfrac{4}{7}\times\dfrac{1}{4}\rightarrow\dfrac{4\times 1}{7\times 4}\rightarrow\dfrac{4}{28}\rightarrow\dfrac{1}{7}$

$\sf\dfrac{4}{7}\times\dfrac{1}{6}\rightarrow\dfrac{4\times 1}{7\times 6}\rightarrow\dfrac{4}{42}\rightarrow\dfrac{2}{21}$

To find out which one is bigger, we need a common denominator. The LCM of 7 and 21 is 21, so convert the first fraction into a denominator of 21. 7 goes into 21 three times, multiply this to the numerator and denominator:

$\sf\dfrac{1}{7}\rightarrow\dfrac{1\times 3}{7 \times 3}\rightarrow\dfrac{3}{21}$

Now just compare the numerators to see which one is bigger.

$\sf\dfrac{3}{21}>\dfrac{2}{21}$ therefore, $\sf\dfrac{4}{7}\times\dfrac{1}{4}>\dfrac{4}{7}\times\dfrac{1}{6}$