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3 years ago

Help please this is so hard I don't understand it

Mathematics

1 answer:

Zina [86]3 years ago

5 0

Answer:

<ABC = 15

< BAC = 150

Step-by-step explanation:

<C = <B since this is an isosceles triangle. We know this because AC = AB

<C = 15 so <ABC = 15

The sum of the angles of a triangle is 180

A + B+ C = 180

A + 15+15 = 180

A +30 =180

A = 180-30

A = 150

< BAC = 150

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Simplify 4(2n -1) – 6n

larisa86 [58]

Answer:

2n - 4

Step-by-step explanation:

expand brackets

8n - 4 - 6n

collect like terms

2n - 4

4 0

3 years ago

Read 2 more answers

What is the value of n in the relationship 8n+9=-n+5

vitfil [10]

Subtract 5 from both sides.

$8n + 4 = -n$

Subtract 8n from both sides.

$4 = -9n$

Divide both sides by -9 to solve for n.

$- \frac{4}{9} = n$

The value of n is -4/9.

8 0

3 years ago

Plsss helppppppp plsssss

MrRa [10]

Set up an equation of ratios as shown below:

h 24 ft 4
----- = ---------- = ---
4 6 ft 1

Cross multiplying the 1st and 3rd fractions, we get h=16 ft (answer)

3 0

3 years ago

PLZ HELP!!! Use limits to evaluate the integral.

Marrrta [24]

Split up the interval [0, 2] into <em>n</em> equally spaced subintervals:

$\left[0,\dfrac2n\right],\left[\dfrac2n,\dfrac4n\right],\left[\dfrac4n,\dfrac6n\right],\ldots,\left[\dfrac{2(n-1)}n,2\right]$

Let's use the right endpoints as our sampling points; they are given by the arithmetic sequence,

$r_i=\dfrac{2i}n$

where $1\le i\le n$ . Each interval has length $\Delta x_i=\frac{2-0}n=\frac2n$ .

At these sampling points, the function takes on values of

$f(r_i)=7{r_i}^3=7\left(\dfrac{2i}n\right)^3=\dfrac{56i^3}{n^3}$

We approximate the integral with the Riemann sum:

$\displaystyle\sum_{i=1}^nf(r_i)\Delta x_i=\frac{112}n\sum_{i=1}^ni^3$

Recall that

$\displaystyle\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}4$

so that the sum reduces to

$\displaystyle\sum_{i=1}^nf(r_i)\Delta x_i=\frac{28n^2(n+1)^2}{n^4}$

Take the limit as <em>n</em> approaches infinity, and the Riemann sum converges to the value of the integral:

$\displaystyle\int_0^27x^3\,\mathrm dx=\lim_{n\to\infty}\frac{28n^2(n+1)^2}{n^4}=\boxed{28}$

Just to check:

$\displaystyle\int_0^27x^3\,\mathrm dx=\frac{7x^4}4\bigg|_0^2=\frac{7\cdot2^4}4=28$

4 0

3 years ago

I am lost on what to do

Neko [114]

$\bf sin({{ \alpha}})sin({{ \beta}})=\cfrac{1}{2}[cos({{ \alpha}}-{{ \beta}})\quad -\quad cos({{ \alpha}}+{{ \beta}})]
\\\\\\
cot(\theta)=\cfrac{cos(\theta)}{sin(\theta)}\\\\
-----------------------------\\\\
\lim\limits_{x\to 0}\ \cfrac{sin(11x)}{cot(5x)}\\\\
-----------------------------\\\\
\cfrac{sin(11x)}{\frac{cos(5x)}{sin(5x)}}\implies \cfrac{sin(11x)}{1}\cdot \cfrac{sin(5x)}{cos(5x)}\implies \cfrac{sin(11x)sin(5x)}{cos(5x)}$

$\bf \cfrac{\frac{cos(11x-5x)-cos(11x+5x)}{2}}{cos(5x)}\implies \cfrac{\frac{cos(6x)-cos(16x)}{2}}{cos(5x)}
\\\\\\
\cfrac{cos(6x)-cos(16x)}{2}\cdot \cfrac{1}{cos(5x)}\implies \cfrac{cos(6x)-cos(16x)}{2cos(5x)}
\\\\\\
\lim\limits_{x\to 0}\ \cfrac{cos(6x)-cos(16x)}{2cos(5x)}\implies \cfrac{1-1}{2\cdot 1}\implies \cfrac{0}{2}\implies 0$

4 0

4 years ago

Help please this is so hard I don't understand it​

Help please this is so hard I don't understand it