Mathematics: What is Jeremys approximate area of the patio

21.11.2021

What is Jeremys approximate area of the patio

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Mathematics

11. Jeremy ran 14 races. Each race was 5 kilometers long. Approximately how many miles did Jeremy run altogether in the races? (1 kilometer 0.62 mile) A 43 miles B. 70 miles C 113 miles D 200 miles

Step-by-step answer

P Answered by PhD

Step-by-step explanation:

Mathematics

Jeremy walked on a treadmill for 45 minutes at a speed of 4.2 miles per hour. Approximately how far did Jeremy walk?

Step-by-step answer

P Answered by PhD

Distance = 3.15 miles

Step-by-step explanation:

Given:

Speed = 4.2 mph

Time = 45 mins = 0.75 hr

Required:

Distance = ?

Solution:

Distance = Speed * Time

Distance = 4.2 * 0.75

Distance = 3.15 miles

Mathematics

11. Jeremy ran 14 races. Each race was 5 kilometers long. Approximately how many miles did Jeremy run altogether in the races? (1 kilometer 0.62 mile) A 43 miles B. 70 miles C 113 miles D 200 miles

Step-by-step answer

P Answered by PhD

Step-by-step explanation:

Mathematics

Jeremy walked on a treadmill for 45 minutes at a speed of 4.2 miles per hour. Approximately how far did Jeremy walk?

Step-by-step answer

P Answered by PhD

Distance = 3.15 miles

Step-by-step explanation:

Given:

Speed = 4.2 mph

Time = 45 mins = 0.75 hr

Required:

Distance = ?

Solution:

Distance = Speed * Time

Distance = 4.2 * 0.75

Distance = 3.15 miles

Chemistry

Blood substitute. as noted in this chapter, blood contains a total concentration of phosphate of approximately 1 mm and typically has a ph of 7.4. you wish to make 100 liters of phosphate buffer with a ph of 7.4 from nah 2 po 4 (molecular weight, 119. 98 g mol 1) and na 2 hpo 4 (molecular weight, 141. 96 g mol 1). how much of each (in grams) do you need? berg, jeremy m.. biochemistry (p. 25). w. h. freeman. kindle edition.

Step-by-step answer

P Answered by Specialist

Mass of NaH₂PO₄ = 4.707 gMass of Na₂HPO₄ = 8.627 g

Explanation:

The equilibrium relevant for this problem is:

H₂PO₄⁻ ↔ HPO₄⁻² + H⁺

The Henderson–Hasselbalch (H-H) equation is needed to solve this problem:

pH= pka + $log\frac{[A^{-} ]}{[HA]}$

In this case, [A⁻] = [HPO₄⁻²], [HA] = [H₂PO₄⁻], pH = 7.4; from literature we know that pka=7.21.

We use the H-H equation to describe [HPO₄⁻²] in terms of [H₂PO₄⁻]:

$7.4=7.21+log\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]} \\0.19=log\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]} \\10^{0.19}= \frac{[HPO4^{-2} ]}{[H2PO4^{-} ]} \\1.549*[H2PO4^{-} ]=[HPO4^{-2} ]$

The problem tells us that the concentration of phosphate is 1 mM, which means:

[HPO₄⁻²] + [H₂PO₄⁻] = 1 mM = 0.001 M

In this equation we can replace [HPO₄⁻²] with the term expressed in the H-H eq:

1.549 * [H₂PO₄⁻] + [H₂PO₄⁻] = 0.001 M

2.549 * [H₂PO₄⁻] = 0.001 M

[H₂PO₄⁻] = 3.923 * 10⁻⁴ M

With the value of [H₂PO₄⁻] we can calculate [HPO₄⁻²]:

[HPO₄⁻²] + 3.923 * 10⁻⁴ M = 0.001 M

[HPO₄⁻²] = 6.077 * 10⁻⁴ M

With the concentrations, the molecular weight, and the volume, we calculate the mass of each reagent:

Mass of NaH₂PO₄ = 3.923 * 10⁻⁴ M * 100 L * 119.98 g/mol = 4.707 gMass of Na₂HPO₄ = 6.077 * 10⁻⁴ M * 100 L * 141.96 g/mol = 8.627 g

Mathematics

4) jeremy bought a new truck for $32,000. the value of the truck after t years can be represented by the formula v = 32,000(.8)t. when will the truck be worth approximately $6700?

Step-by-step answer

P Answered by PhD

V(2) = 32000*0.8^2 = $20,480

Mathematics

Jeremy bought a new truck for $32,000. the value of the truck after t years can be represented by the formula v = 32,000(.8)t. when will the truck be worth approximately $6700?

Step-by-step answer

P Answered by PhD

The truck Jeremy bought new was $32,000. The value of the truck after T years can be represented by the formula V=32,000 (.8)T. The truck will be worth approximately $6,700 in 8 years.

Mathematics

Jeremy bought a new truck for $32,000. the value of the truck after t years can be represented by the formula v = 32,000(.8)^t. when will the truck be worth approximately $6700? a) in 5 years b) in 7 years c) in 8 years d) in 9 years

Step-by-step answer

P Answered by Specialist

divide both sides by 32000
$\frac{67}{320}=0.8^t$
take the ln of both sides
$ln( \frac{67}{320})=ln(0.8^t)$
using the property that ln(a^b)=b(ln(a))

$ln( \frac{67}{320})=t(ln(0.8))$
divide both sides by ln(0.8)
$\frac{ln( \frac{67}{320}) }{ln(0.8)} =t$
use calculator
7.00728=t
so in about 7 years

B

Mathematics

Jeremy bought a new truck for $32,000. the value of the truck after t years can be represented by the formula v = 32,000(.8)t. when will the truck be worth approximately $6700?

Step-by-step answer

P Answered by PhD

7 years

Step-by-step explanation:

Given: The initial price of truck = $32,000

The value of the truck after t years can be represented by the formula ,

V=32,000(0.8)^t

To find the time t in years after which the worth of the truck will be approximately $6700, we need to substitute V=6700 in the equation, we get

$6700=32,000(0.8)^t\\\\\Rightarrow(0.8)^t=\frac{6700}{32000}\\\\\Rightarrow(0.8)^t=0.209375$

Taking log on both sides, we get

$t\log(0.8)=\log(0.209375)\\\\\Rightarrow t=\frac{\log(0.209375)}{\log(0.8)}\\\\\Rightarrow t=7.00727566266\approx7$

Hence, after 7 years the worth of the truck will be approximately $6700.

Chemistry

Step-by-step answer

P Answered by Specialist

Mass of NaH₂PO₄ = 4.707 gMass of Na₂HPO₄ = 8.627 g

Explanation:

The equilibrium relevant for this problem is:

H₂PO₄⁻ ↔ HPO₄⁻² + H⁺

The Henderson–Hasselbalch (H-H) equation is needed to solve this problem:

pH= pka + $log\frac{[A^{-} ]}{[HA]}$

In this case, [A⁻] = [HPO₄⁻²], [HA] = [H₂PO₄⁻], pH = 7.4; from literature we know that pka=7.21.

We use the H-H equation to describe [HPO₄⁻²] in terms of [H₂PO₄⁻]:

$7.4=7.21+log\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]} \\0.19=log\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]} \\10^{0.19}= \frac{[HPO4^{-2} ]}{[H2PO4^{-} ]} \\1.549*[H2PO4^{-} ]=[HPO4^{-2} ]$

The problem tells us that the concentration of phosphate is 1 mM, which means:

[HPO₄⁻²] + [H₂PO₄⁻] = 1 mM = 0.001 M

In this equation we can replace [HPO₄⁻²] with the term expressed in the H-H eq:

1.549 * [H₂PO₄⁻] + [H₂PO₄⁻] = 0.001 M

2.549 * [H₂PO₄⁻] = 0.001 M

[H₂PO₄⁻] = 3.923 * 10⁻⁴ M

With the value of [H₂PO₄⁻] we can calculate [HPO₄⁻²]:

[HPO₄⁻²] + 3.923 * 10⁻⁴ M = 0.001 M

[HPO₄⁻²] = 6.077 * 10⁻⁴ M

With the concentrations, the molecular weight, and the volume, we calculate the mass of each reagent:

Mass of NaH₂PO₄ = 3.923 * 10⁻⁴ M * 100 L * 119.98 g/mol = 4.707 gMass of Na₂HPO₄ = 6.077 * 10⁻⁴ M * 100 L * 141.96 g/mol = 8.627 g

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