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Alex Ar [27]
3 years ago
13

HELP HELP HELP OMGOMG

Mathematics
1 answer:
user100 [1]3 years ago
6 0

Answer:

Direct variation

Step-by-step explanation:

It's not iverse because the line is curved so it's direct variation

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Simplify the following expressions using Order of Operations.<br><br> 4+ 8 = 2 + 6 x 2
Hunter-Best [27]

Answer:

12 ≠ 14

Step-by-step explanation:

(4+8)=2+(6x2)

12 = 2 + 12

12 ≠ 14.

So the equation is false.

4 0
3 years ago
The lengths, in centimeters, of nine earthworms are shown below. 3, 4, 5, 5, 6, 7, 8, 9, 10 What is the median of the data?
mel-nik [20]
The median of the problem is 6 because it is in the middle of the number set of the earthworm sizes.
6 0
3 years ago
100 points , please help. I am not sure if I did this correct if anyone can double-check me thanks!
Nookie1986 [14]

Step-by-step explanation:

\lim_{n \to \infty} \sum\limits_{k=1}^{n}f(x_{k}) \Delta x = \int\limits^a_b {f(x)} \, dx \\where\ \Delta x = \frac{b-a}{n} \ and\ x_{k}=a+\Delta x \times k

In this case we have:

Δx = 3/n

b − a = 3

a = 1

b = 4

So the integral is:

∫₁⁴ √x dx

To evaluate the integral, we write the radical as an exponent.

∫₁⁴ x^½ dx

= ⅔ x^³/₂ + C |₁⁴

= (⅔ 4^³/₂ + C) − (⅔ 1^³/₂ + C)

= ⅔ (8) + C − ⅔ − C

= 14/3

If ∫₁⁴ f(x) dx = e⁴ − e, then:

∫₁⁴ (2f(x) − 1) dx

= 2 ∫₁⁴ f(x) dx − ∫₁⁴ dx

= 2 (e⁴ − e) − (x + C) |₁⁴

= 2e⁴ − 2e − 3

∫ sec²(x/k) dx

k ∫ 1/k sec²(x/k) dx

k tan(x/k) + C

Evaluating between x=0 and x=π/2:

k tan(π/(2k)) + C − (k tan(0) + C)

k tan(π/(2k))

Setting this equal to k:

k tan(π/(2k)) = k

tan(π/(2k)) = 1

π/(2k) = π/4

1/(2k) = 1/4

2k = 4

k = 2

8 0
3 years ago
Solve p(x+q)=r for x
STatiana [176]
(x+q) = r/p
x+q = r/p
x = (r/p) - q
3 0
3 years ago
[4]
Tanya [424]

Answer:

Angle ACD = 38°

Step-by-step explanation:

The full, correct question is presented the attached image to this solution.

Given

Point O is the centre if the circle

Points A, B, C and D are points on the circle

Angle AOB = 140°

Angle OAC = 14°

Angle AOB = 2 × (Angle ACB) [angle subtended at the centre of the circle is twice the angle subtended at the circumference of the circle)

140° = 2 × (Angle ACB)

Angle ACB = (140°/2) = 70°

(Angle AOB) + (Angle OAB) + (Angle ABO) = 180° [sun of angles in a triangle is 180°]

But Angle OAB = Angle ABO = a [base angles of an iscosceles triangle are equal since OA and OB are both radii for the circle O)

(Angle AOB) + (Angle OAB) + (Angle ABO) = 180°

140° + a + a = 180°

2a = 40°

a = (40/2) = 20°

Angle OAB = Angle ABO = 20°

Angle CAB = (Angle OAC) + (Angle OAB) = 14° + 20° = 34°

Triangle ADC is an iscosceles triangle, hence,

Angle DAC = Angle ACD = x [base angles of an iscosceles triangle are equal]

But

(Angle DCB) + (Angle DAB) = 180° [Opposite angles of a cyclic quadilateral (a quadilateral inscribed in a circle) sum up to give 180°]

Angle DCB = (Angle DCA) + (Angle ACB) = (x + 70°)

Angle DAB = (Angle DAC) + (Angle CAB) = (x + 34°)

(Angle DCB) + (Angle DAB) = 180°

(x + 70°) + (x + 34°) = 180°

2x + 104° = 180°

2x = 180° - 104° = 76°

x = (76°/2) = 38°

Angle DAC = Angle ACD = 38°

Angle ACD = 38°

Hope this Helps!!!

7 0
3 years ago
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