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Makovka662 [10]
3 years ago
14

Please help; what’s the measure of angle 1 and 2?

Mathematics
1 answer:
SOVA2 [1]3 years ago
6 0

Answer:

4x-28+x+68=180

5x=180+28-68

5x=140

X=28

D would be the answer

Step-by-step explanation:

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Is (the number in the photo) rational or irrational?
Leno4ka [110]

Answer:

irrational

Step-by-step explanation:

4 0
2 years ago
There are two consecutive even numbers such
Bond [772]

Step-by-step explanation:

Let the numbers be x and y

2 consecutive even nos.

therefore y = x+2

twice of the first (2x) is 46 more than y

2x = y+46

Solving both the equations

x - y + 2 = 0

(-)2x - y - 46 = 0

____________

-x + 0 + 48 = 0

____________

x = 48

y = x+2

= 48 +2

= 50

Therefore the 2 nos. are 48 and 50

Hope this answers helps you

3 0
2 years ago
Will ran four miles on his first day of training. The next day he ran one-third that distance. How far did he run the second day
igomit [66]
1.3333333333333333333333333333333333333
4 0
2 years ago
Can anyone solve part C?​
nlexa [21]

Answer:

y - 7 = 4(x - 35)

Step-by-step explanation:

The fundamental theorem of calculus states that:

\frac{d}{dx} \int\limits^x_a {f(t)} \, dt = f(x).

So using the fundamental theorem of calculus, you can find that h'(x) = f(x).

The question tells you that f(x) is periodic with a period of 8, so f(x) repeats itself every 8 units.

Using this, you can find that the slope of h(x) at x = 35 is the same as the slope of h(x) at x = 3, which is 4.

The slope of h(x) at x = 35 is 4.

Now I have to find the value of h(x) when x = 35. It is the area under f(x) from 0 to 35.

The area underneath f(x) from 0 to 35 is 7. When x = 35, h(x) = 7.

Now use the point-slope formula to write the equation of the tangent line.

The answer is <u>y - 7 = 4(x - 35)</u>

3 0
3 years ago
Read 2 more answers
$5 PAYPAL + BRAINLIEST TO ANSWER THS
expeople1 [14]

Answer:

Quadratic Equation:

3x^2=2x +5

\text{Standard Form: } 3x^2-2x-5=0

From the standard form of a Quadratic Function, we get:

a=3,\:b=-2,\:c=-5

Discriminant:

\Delta=\left(-2\right)^2-4\cdot \:3\left(-5\right)

\Delta=\left(-2\right)^2+4\cdot \:3\cdot \:5

\Delta=64

From the discriminant, we conclude that the equation will have two real solutions.

State that:

b^2-4ac

b^2-4ac =0:\text{The equation has 1 real solution}

b^2-4ac >0:\text{The equation has 2 real solutions}

By the way, solving the equation given:

$x=\frac{2\pm\sqrt{64}}{2\cdot \:3}$

$x=\frac{2\pm\sqrt{64}}{6}$

$x=\frac{2\pm8}{6}$

$x_{1} =\frac{10}{6}=\frac{5}{3}  $

$x_{2}=\frac{-6}{6} =-1$

5 0
3 years ago
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