Please help me im stuck

Mathematics

1 answer:

dmitriy555 [2]3 years ago

6 0

Answer:

a = -0.5

Step-by-step explanation:

$\rm Solve \: for \: a: \\ \rm \longrightarrow - \dfrac{1}{4}a - 4 = \dfrac{7}{4}a - 3 \\ \\ \rm Put \: each \: term \: in \: - \dfrac{1}{4}a - 4 \: over \: the \\ \rm common \: denominator \: 4: \\ \rm - \dfrac{a}{4} - 4 = - \dfrac{a}{4} - \dfrac{16}{4} : \\ \rm \longrightarrow - \dfrac{a}{4} - \dfrac{16}{4} = \dfrac{7}{4}a - 3 \\ \\ \rm - \dfrac{a}{4} - \dfrac{16}{4} = \dfrac{ - a - 16}{4} : \\ \rm \longrightarrow \dfrac{ - a - 16}{4} = \dfrac{7}{4}a - 3 \\ \\ \rm Put \: each \: term \: in \: \dfrac{7}{4}a - 3 \: over \: the \\ \rm common \: denominator \: 4: \\ \rm \dfrac{7a}{4} - 3 = \dfrac{7a}{4} - \dfrac{12}{4} : \\ \rm \longrightarrow \dfrac{ - a - 16}{4} = \dfrac{7a}{4} - \dfrac{12}{4} \\ \\ \rm \dfrac{7a}{4} - \dfrac{12}{4} = \frac{7a - 12}{4} : \\ \rm \longrightarrow \dfrac{ - a - 16}{4} = \dfrac{7a - 12}{4} \\ \\ \rm Multiply \: both \: sides \: by \: 4: \\ \rm \longrightarrow \dfrac{ - a - 16}{ \cancel{4}} \times \cancel{4}= \dfrac{7a - 12}{ \cancel{4}} \times \cancel{4} \\ \\ \rm \longrightarrow -a - 16 = 7 a - 12 \\ \\ \rm Subtract \: 7 a \: from \: both \: sides: \\ \rm \longrightarrow (-a - 7 a) - 16 = (7 a - 7 a) - 12 \\ \\ \rm -a - 7 a = -8 a: \\ \rm \longrightarrow -8 a - 16 = (7 a - 7 a) - 12 \\ \\ \rm 7 a - 7 a = 0: \\ \rm \longrightarrow -8 a - 16 = -12 \\ \\ \rm Add \: 16 \: to \: both \: sides: \\ \rm \longrightarrow (16 - 16) - 8 a = 16 - 12 \\ \\ \rm 16 - 16 = 0: \\ \rm \longrightarrow -8 a = 16 - 12 \\ \\ \rm 16 - 12 = 4: \\ \rm \longrightarrow -8 a = 4 \\ \\ \rm Divide \: both \: sides \: of \: -8 a = 4 \: by \: -8: \\ \rm \longrightarrow \dfrac{ - 8a}{ - 8} = \dfrac{4}{ - 8} \\ \\ \rm \dfrac{ - 8}{ - 8} = 1: \\ \rm \longrightarrow a = - \dfrac{4}{ 8} \\ \\ \rm - \dfrac{4}{ 8} = - \dfrac{1}{2} : \\ \rm \longrightarrow a = - \dfrac{1}{2} \\ \\ \rm \longrightarrow a = - 0.5$

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Elena bikes 20 minutes each day for exercise. Write an equation to describe the relationship between her distance in miles, D, a

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Answer:

The equation to describe the relationship between Elena distance:

a) D = 4.34 miles, b) D = 4.25 miles and c) 12D = M + 3N miles.

Solution:

Given, Elena bikes 20 minutes each day for exercise.

We have to write an equation to describe the relationship between her distance in miles, D, and her biking speed, in miles per hour,

We know that, distance travelled = speed $\times$ time

a. At a constant speed of 13 miles per hour for the entire 20 minutes

Her speed is 13 miles per hour.

Then, distance D miles = 13 miles per hour $\times$ 20 minutes

$\mathrm{D}=13 \text { miles per hour } \times \frac{20}{60} \text { hours } \rightarrow \mathrm{d}=13 \times \frac{1}{3} \rightarrow \mathrm{d}=4.34 \text { miles approximately. }$

b. At a constant speed of 15 minutes per hour for the first 5 minutes, then at 12 miles per hour for the last 15 minutes

Now, total distance travelled = distance travelled with 15 mph + distance travelled with 12 mph

$\begin{array}{l}{\mathrm{D}=15 \mathrm{mph} \times 5 \text { minutes }+12 \mathrm{mph} \times 15 \text { minutes }} \\\\ {\mathrm{D}=15 \mathrm{mph} \times \frac{5}{60} \text { hours }+12 \mathrm{mph} \times \frac{15}{60} \text { minutes }} \\\\ {\mathrm{D}=15 \times \frac{1}{12}+12 \times \frac{1}{4} \rightarrow \mathrm{D}=\frac{5}{4}+3 \rightarrow \mathrm{D}=3+1.25 \rightarrow \mathrm{D}=4.25 \mathrm{miles}}\end{array}$

c. At a constant speed of M miles per hour for the first 5 minutes, then at N miles per hour for the last 15 minutes

Now, total distance travelled = distance travelled with M mph + distance travelled with N mph

$D = M mph \times 5 minutes + N mph \times 15 minutes$

$\mathrm{D}=\mathrm{M} \mathrm{mph} \times \frac{5}{60} \text { hours }+\mathrm{N} \mathrm{mph} \times \frac{15}{60} \text { minutes }$

$\mathrm{D}=\mathrm{M} \times \frac{1}{12}+\mathrm{N} \times \frac{1}{4} \rightarrow \mathrm{D}=\frac{M}{12}+\frac{N}{4} \rightarrow \mathrm{D}=\frac{M}{12}+\frac{3 N}{12} \rightarrow 12 \mathrm{D}=\mathrm{M}+3 \mathrm{N} \text { miles }$