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2 years ago

In the figure below, points J, K, and L are the midpoints of the sides of AXYZ.

Mathematics

2 answers:

Dima020 [189]2 years ago

8 0

Answer:

28 ; 76 ; 38

Step-by-step explanation:

KL = $\frac{XY}{2}$ = $\frac{56}{2}$ = 28 units

YZ = 2 × JK = 38 × 2 = 76 units

LZ = 76/2 = 38 units

Nataliya [291]2 years ago

8 0

Answer:

38, 76 and 28

Step-by-step explanation:

LZ = 76/2 = 38 units

KL = 28 units

YZ = 2 × JK = 38 × 2 = 76 units

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Find domain of y=rad20-6x

Evgesh-ka [11]

Domain, on an "even root" context, means, an even root cannot have a negative radicand, since say for example $\bf \sqrt{-25}\ne -5\qquad why?\implies (-5)(-5)=25$

so... you end up with an "imaginary value"

so... for the case of 20-6x
"x", the domain, or INPUT
can afford to have any value, so long it doesn't make the radicand negative

to check for that, let us make the expression to 0, and see what is "x" then

$\bf 20-6x=0\implies 20=6x\implies \cfrac{20}{6}=x\implies \cfrac{10}{3}=x$

now, if "x" is 10/3, let's see $\bf \sqrt{20-6\left( \frac{10}{3} \right)}\implies \sqrt{20-20}\implies \sqrt{0}\implies 0$

now, 0 is not negative, so the radicand is golden

BUT, if "x" has a value higher than 10/3, the radicand turns negative
for example $\bf x=\frac{11}{3}
\\\\\\

\sqrt{20-6\left( \frac{11}{3} \right)}\implies \sqrt{20-22}\implies \sqrt{-2}$

so.. that's not a good value for "x"

thus, the domain, or values "x" can safely take on, are all real numbers from 10/3 onwards, or to infinity if you wish

5 0

3 years ago

Which number line can be used to find the distance between (-1, 2) and (-5, 2)

wariber [46]

Answer:

y = 2; distance is 4 units

Step-by-step explanation:

Line y = 2

Because the y- coordinate is the same, only horizontal movement from -1 to -5..

Which is 4 units

6 0

3 years ago

Read 2 more answers

(8y²)(-3x²y²)(2/3xy⁴) HELP PLEASEEEEEEEE

IRINA_888 [86]

Step-by-step explanation:

1 Remove parentheses.

8{y}^{2}\times -3{x}^{2}{y}^{2}\times \frac{2}{3}x{y}^{4}

×−3x

2 Use this rule: \frac{a}{b} \times \frac{c}{d}=\frac{ac}{bd}

\frac{8{y}^{2}\times -3{x}^{2}{y}^{2}\times 2x{y}^{4}}{3}

×−3x

×2xy

3 Take out the constants.

\frac{(8\times -3\times 2){y}^{2}{y}^{2}{y}^{4}{x}^{2}x}{3}

(8×−3×2)y

4 Simplify 8\times -38×−3 to -24−24.

\frac{(-24\times 2){y}^{2}{y}^{2}{y}^{4}{x}^{2}x}{3}

(−24×2)y

5 Simplify -24\times 2−24×2 to -48−48.

\frac{-48{y}^{2}{y}^{2}{y}^{4}{x}^{2}x}{3}

−48y

6 Use Product Rule: {x}^{a}{x}^{b}={x}^{a+b}x

a+b

\frac{-48{y}^{2+2+4}{x}^{2+1}}{3}

−48y

2+2+4

2+1

7 Simplify 2+22+2 to 44.

\frac{-48{y}^{4+4}{x}^{2+1}}{3}

−48y

4+4

2+1

8 Simplify 4+44+4 to 88.

\frac{-48{y}^{8}{x}^{2+1}}{3}

−48y

2+1

9 Simplify 2+12+1 to 33.

\frac{-48{y}^{8}{x}^{3}}{3}

−48y

10 Move the negative sign to the left.

-\frac{48{y}^{8}{x}^{3}}{3}

−

48y

11 Simplify \frac{48{y}^{8}{x}^{3}}{3}

48y

to 16{y}^{8}{x}^{3}16y

-16{y}^{8}{x}^{3}

−16y

Done

6 0

2 years ago

K+(-5)=-4 solve each equation

Otrada [13]

Answer:

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

how do you solve y=x+4 and y=-2x+1 using the subsituation method and show work and please give me the correct answer

SIZIF [17.4K]

Answer:

(-1,3)

Step-by-step explanation:

so y=x+4 and y=-2x+1

obviously, y=y

so that means that x+4 is equal to -2x+1

let's use that

y=x+4,y=-2x+1

y=y

x+4=-2x+1

x+2x=1-4

3x=-3

x=-1

y=x+4

y=(-1)+4

y=3

(-1,3)

6 0

2 years ago