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2 years ago

Solve for x and y when 1x -250y = 86

Mathematics

2 answers:

RSB [31]2 years ago

7 0

Answer:

x=250y+86

Step-by-step explanation:

hope this helps

Elodia [21]2 years ago

4 0

Answer:

x=250y+86

Step-by-step explanation:

add 250 to both sides

x−250y+250y=86+250y

x=250y+86

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Order from least to greatest ! 1.36,1.3,1.63 and this other 0.42,3.74,3.47.

frosja888 [35]

If you are putting them all in order from least to greatest it would be 0.42, 1.3, 1.36, 1.63, 3.47,3.74. But, if you are separating them the first set of three numbers you wrote would be ordered 1.3, 1.36, 1.63. Then, the second set would be ordered 0.42, 3.47, 3.74. Hope this helped!!!

4 0

3 years ago

Read 2 more answers

I've been stuck on this for a long time. I would greatly appreciate help.

choli [55]

The vertex is (8,7)
the minimum value is 7
the vertex form is y=(x-8)^2 +7
the minimum value of y=7

i hope this helps!

7 0

3 years ago

Given the piecewise function shown below, select all of the statments that are true.

timurjin [86]

The given the piecewise function is :

$f(x) = \[ \begin{cases} 
 2x & x \ \textless \ 1 \\
 5 & x=1 \\
 x^2 & x\ \textgreater \ 1 
 \end{cases}
\]$

To find f(5) ⇒ substitute with x = 5 in the function → x²
∴ f(5) = 5² = 25

To find f(2) ⇒ substitute with x = 5 in the function → x²
∴ f(2) = 2² = 4

To find f(-2) ⇒ substitute with x = 5 in the function → 2x
∴ f(-2) = 2 * (-2) = -4

To find f(1) ⇒ substitute with x = 1 in the function → 5
∴ f(1) = 5
================================
So, the statements which are true:
$\framebox {B. f(2) = 4} \ \framebox {D. f(1) = 5}$

8 0

3 years ago

Mark and don are planning to sell each of their marble collections at a garage sale. if don has 4 more than 5 times the number o

natita [175]

Mark and Don have to sell 11 and 59 marbles respectively.

Explanation

Lets assume, the number of marbles Mark has is $x$

As, Don has 4 more than 5 times the number of marbles mark has, so the number of marbles Don has $= 5x+ 4$

Question says that the total number of marbles is 70. So, the equation will be...

$x+(5x+4)= 70\\ \\ 6x+4= 70 \\ \\ 6x = 70-4\\ \\ 6x=66 \\ \\ x= \frac{66}{6}= 11$

So, the number of marbles Mark has = 11

and the number of marbles Don has = $(5*11)+4 = 55+4= 59$

6 0

3 years ago

Which equation has the solutions x = -3 ± √3i/2 ?

Maurinko [17]

Answer:Answer is option C : [ $x^{2}$ + 3x + 3 ] =0

Note: None of options matches with given question.

instead of "-3" , there should be "- $\frac{3}{2}$ ".

Step-by-step explanation:

Note: None of options matches with given question.

instead of "-3" , there should be " $\frac{3}{2}$ ".

Here, First thing you have to observe the nature of roots.

∴ x = - $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i and x = - $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$

∴ [ x+( $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$ i) ][ x+( $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i) ]=0

∴ [ $x^{2}$ + x( $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i)+ x( $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$ i) + ( $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$ i)( $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i) ]=0

∴ [ $x^{2}$ + $\frac{3}{2}$ x + $\frac{\sqrt{3}}{2}$ ix + $\frac{3}{2}$ x - $\frac{\sqrt{3}}{2}$ ix + (3- $\frac{\sqrt{3}}{2}$ i)(3+ $\frac{\sqrt{3}}{2}$ i) ] =0

∴ [ $x^{2}$ + 3x + ( $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$ i)( $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i) ] =0

∴ [ $x^{2}$ + 3x + $\frac{9}{4}$ - ( $\frac{\sqrt{3}}{2}$ i)( $\frac{\sqrt{3}}{2}$ i) ] =0

∴ [ $x^{2}$ + 3x + $\frac{9}{4}$ - ( $\frac{3}{4}$ ) $i^{2}$ ] =0

∴ [ $x^{2}$ + 3x + $\frac{9}{4}$ + ( $\frac{3}{4}$ ) ] =0

∴ [ $x^{2}$ + 3x + $\frac{12}{4}$ ] =0

∴ [ $x^{2}$ + 3x + 3 ] =0

Thus, Answer is option C : [ $x^{2}$ + 3x + 3 ] =0

6 0

3 years ago