Can some one help me with this math problem?

Evaluate a(4b + c2) if a = 2, b = 5, and c = 1.

Maksim231197 [3]

Answer:

42

Step-by-step explanation:

a = 2, b = 5, and c = 1

a*(4b + c²) = ⇒ plug in values
2*(4*5 + 1²) = ⇒ solving exponents
2*(20 + 1) = ⇒ parenthesis
2*21 = ⇒ multiplication
42 ⇒ answer

3 0

3 years ago

Read 2 more answers

Somebody pls help I’m doin a exam!!!!

ra1l [238]

Answer:

Step-by-step explanation:

12x+2y=36

2y=36-12x

5x+36-12x=22

7x=14

x=2

2y=36-12(2)

2y=12

y=6

btw,next time, please do not include words like(failing ,exam,test,etc) in your question,otherwise, the question might be deleted.

5 0

2 years ago

Find 'x' in the following

Juli2301 [7.4K]

Answer:

x = 130

Step-by-step explanation:

Sum of all the angle of quadrilateral = 360°

50 + 45 + 35 + ∠ADC = 360

130 + ∠ADC = 360

∠ADC = 360 - 130

∠ADC = 230

x = Reflex ∠ADC

= 360 - 230

= 130

6 0

3 years ago

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8. Calculate<br> 341) + 35）<br> ༣༨

Marta_Voda [28]

Answer: 341 + 35 = 376

Step-by-step explanation:

5 0

3 years ago

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A person stands 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 meters

In-s [12.5K]

Answer:

Therefore the rate change of distance between the car and the person at the instant, the car is 24 m from the intersection is 12 m/s.

Step-by-step explanation:

Given that,

A person stand 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 m/s.

From Pythagorean Theorem,

(The distance between car and person)²= (The distance of the car from intersection)²+ (The distance of the person from intersection)²+

Assume that the distance of the car from the intersection and from the person be x and y at any time t respectively.

∴y²= x²+10²

$\Rightarrow y=\sqrt{x^2+100}$

Differentiating with respect to t

$\frac{dy}{dt}=\frac{1}{2\sqrt{x^2+100}}. 2x\frac{dx}{dt}$

$\Rightarrow \frac{dy}{dt}=\frac{x}{\sqrt{x^2+100}}. \frac{dx}{dt}$

Since the car driving towards the intersection at 13 m/s.

so, $\frac{dx}{dt}=-13$

$\therefore \frac{dy}{dt}=\frac{x}{\sqrt{x^2+100}}.(-13)$

Now

$\therefore \frac{dy}{dt}|_{x=24}=\frac{24}{\sqrt{24^2+100}}.(-13)$

$=\frac{24\times (-13)}{\sqrt{676}}$

$=\frac{24\times (-13)}{26}$

= -12 m/s

Negative sign denotes the distance between the car and the person decrease.

Therefore the rate change of distance between the car and the person at the instant, the car is 24 m from the intersection is 12 m/s.

8 0

2 years ago