I need help on this math problem.

Mathematics

1 answer:

OLEGan [10]3 years ago

6 0

B: (0, a)

C should have the same y value as B

D: (a,0)

C should have the same x value as D

So point C is (a,a)

Since point A is on the origin, its point is (0,0)

You use the slope formula and plug in point A and C:

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$m=\frac{a-0}{a-0}$

$m=\frac{a}{a}$

m = 1

So the value that belongs in the green box is 1

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Please help ASAP!!

marta [7]

Answer:

D. NM=12; mK=70°

Step-by-step explanation:

KN≈KM so they have the same 55° and then

180°-55°-55°=70° and

6*2=12

3 0

3 years ago

joe weighs 20lbs.less yhantwice jeffs weight.if jeff would gain 10lbs.,then together they wouldweigh 250lbs. how much does each

masya89 [10]

So,

Let x represent Joe's weight and y represent Jeff's weight.

"Joe weighs 20 lbs. less than twice Jeff's weight."
x = 2y - 20

"If Jeff would gain 10 lbs., then together they would weigh 250 lbs."
(y + 10) + x = 250

We now have our two open sentences.
x = 2y - 20
(y + 10) + x = 250

Get rid of parentheses.
x = 2y - 20
x + y + 10 = 250

We will use Elimination by Substitution.
2y - 20 + y + 10 = 250

Collect Like Terms.
3y - 10 = 250

Add 10 to both sides.
3y = 260

Divide both sides by 3.
$Jeff's\ weight = 86 \frac{2}{3} \ lbs.$

Substitute again.
$x = 2(86 \frac{2}{3} )-20$

Multiply.
$x = 173 \frac{1}{3} -20$

Subtract.
$Joe's\ weight = 153 \frac{1}{3}\ lbs.$

Check.

"Joe weighs 20 lbs. less than twice Jeff's weight."
Jeff's weight times two is 173 and one-third.
20 lbs. less than that is 153 and one-third lbs. Check.

"If Jeff would gain 10 lbs., then together they would weigh 250 lbs."
86 and two-thirds + 10 = 96 and two-thirds.
96 and two-thirds + 153 and one-third equals 250 lbs. Check.

$Joe's\ weight\ is\ 153 \frac{1}{3} \ lbs.$
$Jeff's\ weight\ is\ 86 \frac{2}{3} \ lbs.$

3 0

3 years ago

Pls help me I don't understand

frutty [35]

Answer:

$d. \: \: \: 25 {x}^{6}$