If cose =-2/3 -, which of the following are possible for the same value of e?

(y − 1)(y + 10)(y-6) < 0

Aliun [14]

Solve for y by simplifying both sides of the inequality, then isolating the variable.

Inequality Form:

y<−10 or 1<y<6

Interval Notation:

(−∞,−10)∪(1,6)

7 0

1 year ago

Please help it’s urgent

Setler [38]

s is less than or equal to 75

3 0

3 years ago

H(x)= ⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ 8+6x x+4 10 4x−3 , , , x<−7 −7≤x<−5 x≥−5

motikmotik

Answer:

$h(-6) = -\frac{1}{5}$

Step-by-step explanation:

Given

$h(x) =\left[\begin{array}{cc}8 + 6x\ \ &x < -7\\\frac{x + 4}{10} &-7 \le x$

Required

Determine h(-6)

To do this, we make use of:

$h(x) = \frac{x + 4}{10}$

Because $-7 \le -6$

So, we have:

$h(x) = \frac{x + 4}{10}$

$h(-6) = \frac{-6 + 4}{10}$

$h(-6) = \frac{-2}{10}$

Simplify

$h(-6) = -\frac{1}{5}$

4 0

3 years ago

Solve the following using Substitution method 2x – 5y = -13 3x + 4y = 15

Digiron [165]

$\huge \boxed{\mathfrak{Question} \downarrow}$

Solve the following using Substitution method

2x – 5y = -13

3x + 4y = 15

$\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}$

$\left. \begin{array} { l } { 2 x - 5 y = - 13 } \\ { 3 x + 4 y = 15 } \end{array} \right.$

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

$2x-5y=-13, \: 3x+4y=15$

Choose one of the equations and solve it for x by isolating x on the left-hand side of the equal sign. I'm choosing the 1st equation for now.

$2x-5y=-13$

Add 5y to both sides of the equation.

$2x=5y-13$

Divide both sides by 2.

$x=\frac{1}{2}\left(5y-13\right) \\$

Multiply $\frac{1}{2}\\$ times 5y - 13.

$x=\frac{5}{2}y-\frac{13}{2} \\$

Substitute $\frac{5y-13}{2}\\$ for x in the other equation, 3x + 4y = 15.

$3\left(\frac{5}{2}y-\frac{13}{2}\right)+4y=15 \\$

Multiply 3 times $\frac{5y-13}{2}\\$ .

$\frac{15}{2}y-\frac{39}{2}+4y=15 \\$

Add $\frac{15y}{2} \\$ to 4y.

$\frac{23}{2}y-\frac{39}{2}=15 \\$

Add $\frac{39}{2}\\$ to both sides of the equation.

$\frac{23}{2}y=\frac{69}{2} \\$

Divide both sides of the equation by 23/2, which is the same as multiplying both sides by the reciprocal of the fraction.

$\large \underline{ \underline{ \sf \: y=3 }}$

Substitute 3 for y in $x=\frac{5}{2}y-\frac{13}{2}\\$ . Because the resulting equation contains only one variable, you can solve for x directly.

$x=\frac{5}{2}\times 3-\frac{13}{2} \\$

Multiply 5/2 times 3.

$x=\frac{15-13}{2} \\$

Add $-\frac{13}{2}\\$ to $\frac{15}{2}\\$ by finding a common denominator and adding the numerators. Then reduce the fraction to its lowest terms if possible.

$\large\underline{ \underline{ \sf \: x=1 }}$

The system is now solved. The value of x & y will be 1 & 3 respectively.

$\huge\boxed{ \boxed{\bf \: x=1, \: y=3 }}$

8 0

3 years ago

Put the following equation of a line into slope-intercept form, simplifying all fractions.

Westkost [7]

Y=1 1/4(1.25)x +9, add 8x to each side and then divide each side by 6

3 0

3 years ago