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3 years ago

Give the digits in the thousands place and the tens place.5,348thousands:tens:

Mathematics

1 answer:

sleet_krkn [62]3 years ago

3 0

Answer:

thousands : 5

hundreds : 3

tens : 4

ones : 8

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A group of 7 people want to equally share 33 1/4 ounces of cake. How much will one person get

AURORKA [14]

Each person will get 4.75 ounces of cake. You just divide 33 1/4 by 7.
Hope this helps! :D

~PutarPotato

5 0

3 years ago

The equation y = \large 1\frac{1}{2}x represents the number of cups of dried fruit, y, needed to make x pounds of granola. Deter

Natasha2012 [34]

Answer:

$(1\frac{1}{2},1)$ - False

$(4,6)$ - True

$(18,12)$ -- False

$(0,0)$ -- True

$(2\frac{1}{2},3\frac{3}{4})$ -- True

Step-by-step explanation:

The points are

$(1\frac{1}{2},1)$ , $(4,6)$ , $(18,12)$ , $(0,0)$ and $(2\frac{1}{2},3\frac{3}{4})$ ---- missing from the question

Given

$y = 1\frac{1}{2}x$

Required

Determine if each of the points would be on $y = 1\frac{1}{2}x$

To do this, we simply substitute the value of x and of each point in $y = 1\frac{1}{2}x$ .

(a) $(1\frac{1}{2},1)$

In this case;

$x = 1\frac{1}{2}$ and $y = 1$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 1\frac{1}{2}$

$y = \frac{3}{2} * \frac{3}{2}$

$y = \frac{9}{4}$

$y = 2\frac{1}{4}$

The point $(1\frac{1}{2},1)$ won't be on the graph because the corresponding value of y for $x = 1\frac{1}{2}$ is $y = 2\frac{1}{4}$

(b) $(4,6)$

In this case;

$x = 4$

$y = 6$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 4$

$y = \frac{3}{2} * 4$

$y = \frac{3* 4}{2}$

$y = \frac{12}{2}$

$y = 6$

The point $(4,6)$ would be on the graph because the corresponding value of y for $x = 4$ is $y = 6$

(c) $(18,12)$

In this case:

$x = 18;y = 12$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 18$

$y = \frac{3}{2} * 18$

$y = \frac{3* 18}{2}$

$y = \frac{54}{2}$

$y = 27$

The point $(18,12)$ wouldn't be on the graph because the corresponding value of y for $x = 18$ is $y = 12$

(d) $(0,0)$

In this case;

$x =0; y = 0$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 0$

$y = 0$

The point $(0,0)$ would be on the graph because the corresponding value of y for $x = 0$ is $y = 0$

(e) $(2\frac{1}{2},3\frac{3}{4})$

In this case:

$x = 2\frac{1}{2}$ ; $y = 3\frac{3}{4}$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 2\frac{1}{2}$

$y = \frac{3}{2} * \frac{5}{2}$

$y = \frac{15}{4}$

$y = 3\frac{3}{4}$

The point $(2\frac{1}{2},3\frac{3}{4})$ would be on the graph because the corresponding value of y for $x = 2\frac{1}{2}$ is $y = 3\frac{3}{4}$

3 0

3 years ago

Solve equation, check for extraneous solutions

fiasKO [112]

Start with

$1=\dfrac{1}{n^2}+\dfrac{n^2-4n-5}{3n^2}$

We observe that both fractions are not defined if $n=0$ . So, we will assume $n \neq 0$ .

We multiply both numerator and denominator of the first fraction by 3 and we sum the two fractions:

$1=\dfrac{3}{3n^2}+\dfrac{n^2-4n-5}{3n^2} = \dfrac{n^2-4n-2}{3n^2}$

We multiply both sides by $3n^2$ :

$n^2-4n-2=3n^2$

We move everything to one side and solve the quadratic equation:

$2n^2+4n+2=0 \iff 2(n^2+2n+1)=0 \iff 2(n+1)^2=0 \iff n+1=0 \iff n=-1$

We check the solution:

$1=\dfrac{1}{1}+\dfrac{1+4-5}{3} \iff 1 = 1+0$

which is true

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3 years ago

What is the volume of the cylinder below?

MrRissso [65]

I think it would be 175

4 0

3 years ago

I need help with this question.

Kitty [74]

i'm not sure how to solve the first one but the second is 150

4 0

3 years ago

Give the digits in the thousands place and the tens place.5,348thousands:tens:​

Give the digits in the thousands place and the tens place.5,348thousands:tens: