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4 years ago

6th grade fractions...

Mathematics

1 answer:

blsea [12.9K]4 years ago

4 0

The last pair of numbers bracket the point 3/7.

12/32 = 3/8 < 3/7

15/18 = 5/6 > 3/7

_____

3/7 ≈ 0.428571... (repeating), so will be larger than 3/8 = 0.375, 12/32 = 3/8, and 4/10. That is, 3/7 is larger than the largest number of each pair except the last.

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You can always compare two fractions to see which is larger by cross-multiplying. That is a/b vs c/d gives the same result as ad vs bc. (Put the product on the numerator side of the comparison.)

Here, that would mean comparing 12/32 vs 3/7, we would get 12·7 vs 3·32, or 84 vs 96. The fraction on the left is smaller.

Comparing 3/7 vs 15/18, we would get 3·18 vs 7·15, or 54 vs 105. The fraction on the right is larger.

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Mai biked 7 and 1/4 miles today, and Noah biked 3 5/8 miles. How many times the length of Noah's bike ride was Mai's bike ride?

mamaluj [8]

Answer:

The distance cover by Mai's bike is 2 time the distance cover by Noah's bike

Step-by-step explanation:

Given as :

The distance cover by Mai's bike = $d_1$ = 7 $\dfrac{1}{4}$ miles

I.e $d_1$ = $\dfrac{28+1}{4}$ miles

Or, $d_1$ = $\dfrac{29}{4}$ miles

The distance cover by Noah's bike = $d_2$ = 3 $\dfrac{5}{8}$ miles

I.e $d_2$ = $\dfrac{24 + 5}{8}$ miles

Or, $d_2$ = $\dfrac{29}{8}$ miles

let the number of times that Noah's bike ride was Mai's bike ride = n

So, The distance cover by Mai's bike = n × The distance cover by Noah's bike

Or, $d_$ = n × $d_$

Or, $\dfrac{29}{4}$ miles = n × $\dfrac{29}{8}$ miles

Or, n = $\frac{\frac{29}{4}}{\frac{29}{8}}$

∴ n = $\dfrac{8}{4}$

I.e n = 2

Hence The distance cover by Mai's bike is 2 time the distance cover by Noah's bike . Answer

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3 years ago

On average, college graduates spend more money on the lottery than those with only a high school education. True or false?

Anettt [7]

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3 years ago

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Someone plz help me and plz show working out

MAXImum [283]

Answer:

20y

Step-by-step explanation:

Area of rectangle = area of square

=> length * breadth = side²

=> 50y * 8y = side²

=> 400y² = side²

=> (20y)² = side²

=> 20y = side length of square

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3 years ago

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Let A = { x ∈ R : x 2 − 5 x + 4 ≤ 0 } , B = (3 , 5), and C = (3 , 4]. Show that A ∩ B = C . (Recall, you must show two separate

kolbaska11 [484]

Answer:

You can proceed as follows:

Step-by-step explanation:

First solve the quadratic inequality $x^{2}-5x+4\leq 0$ . To do that, factorize, then we have that $(x-4)(x-1)\leq 0$ . This implies that

$x-1\leq 0\, \text{and}\, x-4\geq 0$

$x-1 \geq 0\, \text{and}\, x-4\leq 0$

In the first case the solution is the empty set $\emptyset$ . In the second case the solution is the interval $1\leq x \leq 4$ . Now we have that

$A=[1,4]$

$B=(3,5)$

$C=(3,4]$ .

To show that $A\cap B\subseteq C$ consider $x\in A\cap B$ . Then $1\leq x \leq 4\, \text{and}\, 1$ , this implies that $3$ , then $x\in C$ . Now, to show that $C\subseteq A\cap B$ consider $x\in C$ , then $3$ , then $1\leq x \leq 4\, \text{and}\, 3$ , then $x\in [1,4] \, \text{and}\, x\in (3,5)$ , this implies that $x\in A\cap B$ .