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3 years ago

How to graph x+y=1 & is it linear

Mathematics

1 answer:

liq [111]3 years ago

7 0

Answer:

Step-by-step explanation:

Plot the points x-intercept and intercept on the graph. Join the two points to get the straight line. This is the graph of the linear equation.

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if mike invests 2000 in the bank at a simple interest rate of 5.3% for 10 years how much interest will he make?

Tems11 [23]

Answer:

$1060

Step-by-step explanation:

First, converting R percent to r a decimal

r = R/100 = 5.3%/100 = 0.053 per year,

then, solving our equation

I = 2000 × 0.053 × 10 = 1060

I = $ 1,060.00

The simple interest accumulated

on a principal of $ 2,000.00

at a rate of 5.3% per year

for 10 years is $ 1,060.00.

6 0

2 years ago

John ran the 100-m dash with a time of 9.81 sec. if this pace could be maintained for an entire 26-mi marathon, what would his t

Dmitry_Shevchenko [17]

Answer:

4104.79 Seconds or about 1.14 Hours

Step-by-step explanation:

26mi = 41842.9m

100/9.81 = 41842.9/x

41842.9 x 9.81 = 100x

410478.849 = 100x

x = 4104.78849

7 0

3 years ago

If you could please kindly help :D

BartSMP [9]

Answer:

y = -1/2 ( x − 4 )^2 + 8

Step-by-step explanation:

First find the vertex form

y = a ( x − h )^ 2 + k where ( h,k) is the vertex

The vertex is ( 4,8)

y = a ( x − 4 )^2 + 8

Now we need to determine a

Pick a point on the graph (0,0)

and substitute this point into the equation

0 = a ( 0 − 4 )^2 + 8

0 = a (16) +8

-8 = 16a

-8/16 =a

-1/2

y = -1/2 ( x − 4 )^2 + 8

7 0

3 years ago

How much heat in kJ is released when nitrogen and hydrogen react as below to form 100.g ammonia heat equals-92.2kj

tresset_1 [31]

100g of ammonia releases 271.18 kJ of heat.

<u>Explanation:</u>

Given:

N₂ reacts with H₂

The reaction is:

2N₂ + 3H₂ → 2NH₃ + 92.2 kJ

According to the balanced reaction:

2 X 17g of ammonia releases 92.2 kJ of heat

34g of ammonia releases 92.2 kJ of heat

100g of ammonia will release = $\frac{92.2}{34} X 100$ kJ of heat

= 271.18 kJ

Therefore, 100g of ammonia releases 271.18 kJ of heat.

8 0

3 years ago

A hot air balloon is flying at a constant speed of 20 mi/h at a bearing of N 36° E. There is a 10mi/h cross wind blowing due eas

Ksju [112]

Answer:

The actual speed = 27.12 mi/h

Direction = 36.5° in NE(north of east)

Step-by-step explanation:

As given , A hot air balloon is flying at a constant speed of 20 mi/h at a bearing of N 36° E.

⇒θ = 36°

Let v₀ be the constant speed, then v₀ = 20

Let vₓ be the speed in East direction

$v_{y}$ be the speed in North direction

So,

vₓ = v₀ sin(θ) = 20 sin(36°) + 10 ( As given, There is a 10mi/h cross wind blowing due east.)

⇒vₓ = 20(0.588) + 10 = 11.76 + 10 = 21.76 mi/h

and $v_{y}$ = v₀ cos(θ) = 20 cos(36°) = 20(0.809) = 16.18 mi/h

Now,

the actual speed = √(vₓ)² + ( $v_{y}$ )²

= √(21.76)² + (16.18)²

= √473.498 + 261.792

= √735.29 = 27.12

⇒The actual speed = 27.12 mi/h

Now,

Direction = θ = $tan^{-1}(\frac{v_{y} }{v_{x} } ) = tan^{-1}(\frac{16.18 }{21.76 } ) = tan^{-1}(0.74) = 36.5$

⇒ Direction = 36.5° in NE(north of east)

7 0

3 years ago