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2 years ago

It takes Niamh 3 hours to drive from Ashdown to Bridgeton at an average speed of 50 mph.

Mathematics

1 answer:

Nadya [2.5K]2 years ago

6 0

Answer:

46.7

Step-by-step explanation:

3 × 50 = 150 miles

60 ÷ 40 = 1.5 h.

(150 + 60) ÷ (1.5 + 3) = 46.7 mph

(Speed × Time = Distance)

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What is the measure of E

Verizon [17]

Answer:

The measure of angle E is:

m∠E = 80°

Step-by-step explanation:

Given

The triangle ΔDEF

It is stated that DE=EF and G is the midpoint of EF.

It means the midpoint G has converted the triangle into two equal right-angles triangles ΔDEG and ΔDFG with the right-angle at G.

Given

m∠GDE = 10°

As the right-angle triangle, ΔDEG lies at the right-angle G.

So, m∠DGE = 90°

m∠GDE = 10°

m∠DGE = 90°

m∠E = ?

We know that the sum of angles of a triangle is 180°.

m∠GDE + m∠DGE + m∠E = 180°

10° + 90° + m∠E = 180°

100 + m∠E = 180°

m∠E = 180° - 100

m∠E = 80°

Therefore,

The measure of angle E is: m∠E = 80°

3 0

2 years ago

The integer −282 could represent which three situations?A) owing $282B) depositing $282C) losing 282 pointsD) growing 282 feetE)

Shkiper50 [21]

A, C, and E
I think that it makes sense

7 0

2 years ago

A car is going 50 miles per hour. A rate that describes this would have units of

Over [174]

Answer:

mph or miles per hour

Step-by-step explanation:

5 0

2 years ago

For the given term, find the binomial raised to the power, whose expansion it came from: 15(5)^2 (-1/2 x) ^4

Elina [12.6K]

Answer:

C. $(5-\frac{1}{2})^6$

Step-by-step explanation:

Given

$15(5)^2(-\frac{1}{2})^4$

Required

Determine which binomial expansion it came from

The first step is to add the powers of he expression in brackets;

$Sum = 2 + 4$

$Sum = 6$

Each term of a binomial expansion are always of the form:

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

Where n = the sum above

$n = 6$

Compare $15(5)^2(-\frac{1}{2})^4$ to the above general form of binomial expansion

$(a+b)^n = ......+15(5)^2(-\frac{1}{2})^4+.......$

Substitute 6 for n

$(a+b)^6 = ......+15(5)^2(-\frac{1}{2})^4+.......$

[Next is to solve for a and b]

From the above expression, the power of (5) is 2

Express 2 as 6 - 4

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

By direct comparison of

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

and

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

We have;

$^nC_ra^{n-r}b^r= 15(5)^{6-4}(-\frac{1}{2})^4$

Further comparison gives

$^nC_r = 15$

$a^{n-r} =(5)^{6-4}$

$b^r= (-\frac{1}{2})^4$

[Solving for a]

By direct comparison of $a^{n-r} =(5)^{6-4}$

$a = 5$

$n = 6$

$r = 4$

[Solving for b]

By direct comparison of $b^r= (-\frac{1}{2})^4$

$r = 4$

$b = \frac{-1}{2}$

Substitute values for a, b, n and r in

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

$(5+\frac{-1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

Solve for $^6C_4$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{(6-4)!4!)}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{2!!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5*4!}{2*1*!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5}{2*1}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{30}{2}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^2(\frac{-1}{2})^4+.......$

Check the list of options for the expression on the left hand side

The correct answer is $(5-\frac{1}{2})^6$

3 0

3 years ago

Which choice represents the number of cookies Rhianna must sell for 35 cents each to make more than $15. A) 0.035x > 15 B) 0.

Tanya [424]

0.35x > 15 <=== ur inequality

7 0

2 years ago

Read 2 more answers