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3 years ago

Please help! Will mark brainliest

Mathematics

1 answer:

ser-zykov [4K]3 years ago

5 0

Answer:

Step-by-step explanation:

t1 = a1

t2 = a1 + d

t3 = a1 + 2d

The sum is t1 + t2 + t3 = a1 + a1 + d + a1 + 2d = 3a1 + 3d

Second question

t1 = a1

t2 = a1 + d

t3 = a1 + 2d

t4 = a1 + 3d

...

tn = a1 + (n - 1)*d

Sum = (a1 + a1 + (n- 1)*d) * n / 2

Sum = (2a1 + nd - d) * n / 2

Sum = (2a1*n + n^2*d - nd) / 2

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14+{-2+3[1+3(-6-2)]}

Marrrta [24]

14 + {-2 + 3[1 + 3(-6-2)]}
14 + { -2 + 3[1 + 3(-8)]}
14 + { -2 + 3[1 - 24]}
14 + { -2 + 3[-23]}
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14 + {-71}
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3 0

3 years ago

Read 2 more answers

Can someone help.

r-ruslan [8.4K]

Answer:

m∠C = 38° (nearest whole number).

Step-by-step explanation:

To find the missing angle, use the Sine Rule.

<u>Sine Rule</u> (for angles)

$\sf \dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}$

where:

A, B and C are the angles
a, b and c are the sides opposite the angles

Given:

A = 53°
a = 18
c = 14

To find m∠C, substitute the given values into the formula and solve for C:

$\implies \sf \dfrac{\sin 53^{\circ}}{18}=\dfrac{\sin C}{14}$

$\implies \sf \sin C= \dfrac{14 \sin 53^{\circ}}{18}$

$\implies \sf C= \sin^{-1} \left( \dfrac{14 \sin 53^{\circ}}{18}\right)$

$\implies \sf C= 38.40096301...^{\circ}$

Therefore, m∠C = 38° (nearest whole number).

Learn more about the sine rule here:

brainly.com/question/27089170

brainly.com/question/27704834

5 0

2 years ago

What is the answer to this equation 6x+3-6x=3

aev [14]

Answer:

0=0

Step-by-step explanation:

6x-6x=0

cancel the terms on both sides of the equation

0=0 the sum of two opposites equals zero

this statement is correct

5 0

3 years ago

If a tank holds 5000 gallons of water, which drains from the bottom of the tank in 40 minutes, then Torricelli's Law gives the v

pochemuha

Answer:

V'(t) = $-250(1 - \frac{1}{40}t)$

If we know the time, we can plug in the value for "t" in the above derivative and find how much water drained for the given point of t.

Step-by-step explanation:

Given:

V = $5000(1 - \frac{1}{40}t )^2$ , where 0≤t≤40.

Here we have to find the derivative with respect to "t"

We have to use the chain rule to find the derivative.

V'(t) = $2(5000)(1 - \frac{1}{40} t)d/dt (1 - \frac{1}{40}t )$

V'(t) = $2(5000)(1 - \frac{1}{40} t)(-\frac{1}{40} )$

When we simplify the above, we get

V'(t) = $-250(1 - \frac{1}{40}t)$

If we know the time, we can plug in the value for "t" and find how much water drained for the given point of t.

4 0

4 years ago

Solve the equation using inverse operations. Check your solutions. In your final answer, include all of your work.

Bess [88]

-2x^2=-15-5
-2x^2=-10
x^2=-5
X≠R i am not quite sure but...hope it helps :)

6 0

3 years ago

Read 2 more answers