Passing through (-2,8) and perpendicular to y=2x+5

Diana averages 6 mph walking

Diano4ka-milaya [45]

Answer:

Distance between school and home = 105 miles.

Step-by-step explanation:

Let distance = D miles

Time to take walking/jogging, t1 = D/6 hours

Time by car, t2 = D/35 hours

Average time, t

= (t1+t2)/2

= D(1/6+1/35)

= (D/210)(35+6)

=41D/210 hours

Given t = t1-3 hours

41D/210 - D/6 = 3

Solve for D

41D/210 - 35D/210 = 3

(41-35)D/210 = 3

6D/210 = 3

D = 3*210/6 = 105 miles.

(Appreciate very much Diana's efforts to walk/job 105 miles to school, and we understand how lucky we are in most of Europe and North America)

3 0

3 years ago

Devon drove at the same speed for 4 hours. He drove a total of 120 miles,

ololo11 [35]

He drove 30 miles in 1 hour

3 0

3 years ago

Read 2 more answers

(Wil give Brainliest) Please Helpp

I am Lyosha [343]

Answer:

41< (or equal to) x < (or equal to) 45

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Solve the triangle using the Law of Sines. (Assume a = 7.1, b = 2.9, and ∠A = 70°. Round the length to one decimal place and the

posledela

Answer:

sides

a=7 (rounded)

b=3 (rounded)

c=8 (rounded)

angles

A=70

B=22.57

C=87.43

Step-by-step explanation:

a . given

b . given

c = a·sin(C)/sin(A)=7.54806

A . given

B = arcsin (b·sin(A)/a) = = 0.39393 rad = 22.57° = 22°34'13"

C = 180° - A - B = 1.52593 rad = 87.43° = 87°25'47"

7 0

3 years ago

Sarah is carrying out a series of experiments which involve using mcreasing amounts of a chemical. In the

Marianna [84]

Sarah is carrying out a series of experiments which involve using increasing amounts of a chemical. In the first experiment she uses 6g of the chemical and in the second experiment she uses 7.8 g of the chemical

(i)Given that the amounts of the chemical used form an arithmetic progression find the total amount of chemical used in the first 30 experiments

(ii)Instead it is given that the amounts of the chemical used for a geometric progression. Sarah has a total of 1800 g of the chemical available. Show that the greatest number of experiments possible satisfies the inequality: $1.3^N \leq 91$ and use logarithms to calculate the value of N.

Answer:

(a)963 grams

(b)N=17

Step-by-step explanation:

(a)

In the first experiment, Sarah uses 6g of the chemical

In the second experiment, Sarah uses 7.8g of the chemical

If this forms an arithmetic progression:

First term, a =6g

Common difference. d= 7.8 -6 =1.8 g

Therefore:

Total Amount of chemical used in the first 30 experiments

$S_n=\dfrac{n}{2}[2a+(n-1)d] \\S_{30}=\dfrac{30}{2}[2*6+(30-1)1.8] \\=15[12+29*1.8]\\=15[12+52.2]\\=15*64.2\\=963$ grams$

Sarah uses 963 grams in the first 30 experiments.

(b) If the increase is geometric

First Term, a=6g

Common ratio, r =7.8/6 =1.3

Sarah has a total of 1800 g

Therefore:

Sum of a geometric sequence

$S_n=\dfrac{a(r^N-1)}{r-1} \\1800=\dfrac{6(1.3^N-1)}{1.3-1} \\1800=\dfrac{6(1.3^N-1)}{0.3}\\$Cross multiply\\1800*0.3=6(1.3^N-1)\\6(1.3^N-1)=540\\1.3^N-1=540\div 6\\1.3^N-1=90\\1.3^N=90+1\\1.3^N=91$

Therefore, the greatest possible number of experiments satisfies the inequality

$1.3^N \leq 91$

Next, we solve for N

Changing $1.3^N \leq 91$ to logarithm form, we obtain:

$N \leq log_{1.3}91\\N \leq \dfrac{log 91}{log 1.3}\\ N \leq 17.19$

Therefore, the number of possible experiments, N=17

3 0

3 years ago