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faltersainse [42]

3 years ago

13

Y = square root of x + 6

2 answers:

wariber [46]3 years ago

5 0

Answer:

Step-by-step explanation:

$x^{2} + 36$

ozzi3 years ago

5 0

Step-by-step explanation:

$y = \sqrt{x + 6} \\ y = x ^{2} + 6 { }^{2} \\ y = x { }^{2} + 36$

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Simplify by combining like terms. <br> 12+5t−4+15t+1

quester [9]

Answer:

9+20t

Step-by-step explanation:

Look at the terms and see which has t and which is a number when you see like terms, add them or subtract them together to get like terms.

5 0

3 years ago

Endpoint: (1,7), midpoint: (-10,10)

algol [13]

The answer is (-21, 13) for The second endpoint.

Let's start by calling the known endpoint L and the unknown K. We'll call the midpoint M. In order to find this, we must first note that to find a midpoint we need to take the average of the endpoints. To do this we add them together and then divide by 2. So, using that, we can write a formula and solve for each part of the k coordinates. We'll start with just x values.

(Kx + Lx)/2 = Mx

(Kx + 1)/2 = -10

Kx + 1 = -20

Kx = -21

And now we do the same thing for y values

(Ky + Ly)/2 = My

(Ky + 7)/2 = 10

Ky + 7 = 20

Ky = 13

This gives us the final point of (-21, 13)

5 0

3 years ago

afamily had dinner in a restaurant. If the cost of the cdinner was Rs. 2500, how much did the family pay with 10% service charge

yan [13]

Step-by-step explanation:

Find 10%:

100% = 2500

1% = 2500 divide 100

= 25

10% = 25 x 10 = 250

Find the 13% vat:

13% = 25 x 13 = 325

Find the total amount he got charged:

325 + 250 = 575

Total amount he had to pay for dinner + charges:

2500 + 575 = 3075

Answer: Rs. 3075

3 0

3 years ago

BD is the angle bisector of

AlexFokin [52]

Note: Let us consider, we need to find the $m\angle ABC$ and $m\angle DBC$ .

Given:

In the given figure, BD is the angle bisector of ABC.

To find:

The $m\angle ABC$ and $m\angle DBC$ .

Solution:

BD is the angle bisector of ABC. So,

$m\angle ABD=m\angle DBC$

$3x=x+20$

$3x-x=20$

$2x=20$

Divide both sides by 2.

$x=\dfrac{20}{2}$

$x=10$

Now,

$m\angle DBC=(x+20)^\circ$

$m\angle DBC=(10+20)^\circ$

$m\angle DBC=30^\circ$

And,

$m\angle ABC=(3x)^\circ+(x+20)^\circ$

$m\angle ABC=(4x+20)^\circ$

$m\angle ABC=(4(10)+20)^\circ$

$m\angle ABC=(40+20)^\circ$

$m\angle ABC=60^\circ$

Therefore, $m\angle DBC=30^\circ,m\angle ABD=30^\circ$ and $m\angle ABC=60^\circ$ .

8 0

2 years ago

Helppppppppppppppppp asappppppp

Pani-rosa [81]

Answer:

140

Step-by-step explanation:

<u>As per picture:</u>

∠PTQ and ∠RTS are vertical angles and they are equal

∠PTQ = (x + 28)°
∠RTS = (2x + 16)°
∠PTQ = ∠RTS
x + 28 = 2x + 16
2x - x = 28 - 16
x = 12

∠PTR is supplementary with ∠PTQ and their sum is 180°

∠PTQ = 12 + 28 = 40°
∠PTR = 180 - 40 = 140°

6 0

3 years ago