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3 years ago

Can someone answer 33 please?!

Mathematics

2 answers:

Temka [501]3 years ago

8 0

(x)+(x+1)+(x+2)=228.

3x+3=228
3x=225
x=75
Integer 1: 75
Integer 2: 76
Integer 3: 77

75+76+77=228

kifflom [539]3 years ago

8 0

The integers will be:
x, x+1, x+2
Add them together:
3x + 3 = 228
Subtract 3 from both sides.
3x = 225
Divide both sides by 3
x = 75
x + 1 = 76
x + 2 = 77
CHECK
75 + 76 + 77 = 228

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If anyone knows about definite integrals for calculus then please I request help! I

kicyunya [14]

Answer:

$\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)$

General Formulas and Concepts:

Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Integration

Integrals

Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

U-Substitution

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx$

Step 2: Integrate Pt. 1

Identify variables for u-substitution.

Set u: $\displaystyle u = 4x^{-2}$
[u] Differentiate [Basic Power Rule, Derivative Properties]: $\displaystyle du = \frac{-8}{x^3} \ dx$
[Bounds] Switch: $\displaystyle \left \{ {{x = 9 ,\ u = 4(9)^{-2} = \frac{4}{81}} \atop {x = 5 ,\ u = 4(5)^{-2} = \frac{4}{25}}} \right.$

Step 3: Integrate Pt. 2

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^9_5 {\frac{-8}{x^3}e^\big{4x^{-2}}} \, dx$
[Integral] U-Substitution: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^{\frac{4}{81}}_{\frac{4}{25}} {e^\big{u}} \, du$
[Integral] Exponential Integration: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}(e^\big{u}) \bigg| \limits^{\frac{4}{81}}_{\frac{4}{25}}$
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8} \bigg( e^\Big{\frac{4}{81}} - e^\Big{\frac{4}{25}} \bigg)$
Simplify: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

4 0

3 years ago

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il63 [147K]

Answer:

i’m pretty sure it’s because they make 180°

Step-by-step explanation:

6 0

3 years ago

The equation of a circle is (x + 6)^2 + (y - 4)^2 = 16. The point (-6, 8) is on the circle.

pantera1 [17]

Answer:

y = 8 is the equation of tangent.

Step-by-step explanation:

The equation of the tangent to the circle at (-6,8) is of the form:

y = mx + c

where m is the slope of the tangent and c is the y-intercept.

The point (-6,8) lies on the circle and the tangent line as well.

Hence (-6,8) satisfies the line equation:

8 = m(-6) + c ⇒ c-6m = 8 -------------1

We know that slope of two perpendicular lines are related as:

$m_{1}\times m_{2}=-1$

At any point on the circle, the normal line at a point is always perpendicular to the tangent line at that point.

Hence :

$m_{normal} \times m_{tangent}=-1$

We can find the slope of the normal at point (-6,8) as it passes through the centre of the circle (-6,4) by using the two-points formula for slope.

$m=\frac{y_2-y_1}{x_2-x_1}$

$=\frac{8-4}{-6+6}$

= ∞

Slope of the normal is infinity and hence slope of tangent is -1/∞ = 0

Hence m=0

Putting m=0 in equation 1 we get:

c = 8

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y = 8

6 0

3 years ago

Determine the first four terms of the sequence in which the nth term is...

Anon25 [30]

Answer:

1/5, 1/6, 1/7, 1/8

Step-by-step explanation:

The formula for the sequence is (n+3)!/ (n+4)!

The first terms uses n=1

a1 = (1+3)!/ (1+4)! = 4!/5! = (4*3*2*1)/(5*4*3*2*1) = 1/5

The first terms uses n=2

a2 = (2+3)!/ (2+4)! = 5!/6! = (5*4*3*2*1)/(6*5*4*3*2*1) = 1/6

The first terms uses n=3

a3 = (3+3)!/ (3+4)! = 6!/7! = (6*5*4*3*2*1)/(7*6*5*4*3*2*1) = 1/7

The first terms uses n=4

a4 = (4+3)!/ (4+4)! = 7!/8! = (7*6*5*4*3*2*1)/(8*7*6*5*4*3*2*1) = 1/8

7 0

3 years ago

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maw [93]

I’m pretty sure it’s it’s between f and g. Sorry if I’m not exact but hope this helps

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3 years ago