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2 years ago

Alice says that 0.145 rounds to 0.15 and Roger says that the number rounds to 0.1. explain how both can be correct?

1 answer:

4 0

Because the 5 in the thousanths place rounds the 4 in the hundreds place... and 0.145 can round down because the hundreds place is less than 5... it all depends on ehat place value your rounding to

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What method would you choose to solve the equation 2x^2-7=9? explain why you chose this method.

Nikolay [14]

Answer:

2.87

Step-by-step explanation:

2x^2-7=9

2x^2=9+7

2x^2/2=16/2

(square root)x^2=(square root)8

x=2.87

I used this method because I don't know any other method.

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The steps to derive the quadratic formula are shown below: Step 1 ax2 + bx + c = 0 Step 2 ax2 + bx = − c Step 3 x2 + b over a ti

Nostrana [21]

Answer:

$x + \frac{b}{2a} = \frac{+/ - \sqrt{b^2-4ac} }{2a}$

Step-by-step explanation:

Step 1:

$ax^2+bx+c = 0$

Step 2:

$ax^2+bx = -c$

Step 3:

$\frac{ax^2+bx}{a} = \frac{-c}{a}$

Step 4:

Adding $\frac{b^2}{4a^2}$ to both sides to complete the square

$x^2 + \frac{bx}{a} + \frac{b^2}{4a^2} = \frac{-c}{a} + \frac{b^2}{4a^2}$

Step 5:

$x^2 + \frac{bx}{a} + \frac{b^2}{4a^2} = \frac{-4ac+b^2}{4a^2}$

Step 6:

Taking square root on both sides

$x + \frac{b}{2a} = \frac{+/ - \sqrt{b^2-4ac} }{2a}$

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2 years ago

The perimeter of a triangle is 30 inches. After a dilation, the perimeter is 6 inches. What is the scale factor of the perimeter

Black_prince [1.1K]

1/5 is the answer brainiest plzzz i hope it helps

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2 years ago

. For each of these intervals, list all its elements or explain why it is empty. a) [a, a] b) [a, a) c) (a, a] d) (a, a) e) (a,

Eva8 [605]

Answer:

Elements are of the form

(i) $[a,a]=\{[x,y] : a\leq x\leq a, a\leq y\leq a; a\in \mathbb R\}$

(ii) $[a,b)=\{[x,y) :a\leq x$

(iii) $(a,a]=\{(x,y] :a$

(iv) $(a,a)=\{(x,y): a$

(v) (a,b) where a>b= $\{(x,y) : a>x>b,a>y>b;a>b,a,b \in \mathbb R\}$

(vi) [a,b] where a>b= $\{[x,y] : a\geq x\geq b,a\geq y\geq b;a>b,a,b \in \mathbb R\}$

Step-by-step explanation:

Given intervals are,

(i) [a,a] (ii) [a,a) (iii) (a,a] (iv) (a,a) (v) (a,b) where a>b (vi) [a,b] where a>b.

To show all its elements,

(i) [a,a]

Imply the set including aa from left as well as right side.

Its elements are of the form.

$\{[a,a] : a\in \mathbb R\}=\{[0,0],[1, 1],[-1,-1],[2,2],[-2,-2],[3,3],[-3,-3],........\}$

Since there is a singleton element a of real numbers, this set is empty.

Because there is no increment so if $a\in \mathbb R$ then the set [a,a] represents singleton sets, and singleton sets are empty so is [a,a].

(ii) [a,a)

This means given interval containing a by left and exclude a by right.

Its elements are of the form.

$[ 1, 1),[-1,-1),[2,2),[-2,-2),[3,3),[-3,-3),........$

Since there is a singleton element a of real numbers withis the set, this set is empty.

Because there is no increment so if $a\in \mathbb R$ then the set [a,a) represents singleton sets, and singleton sets are empty so is [a.a).

(iii) (a,a]

It means the interval not taking a by left and include a by right.

Its elements are of the form.

$( 1, 1],(-1,-1],(2,2],(-2,-2],(3,3],(-3,-3],........$

Since there is a singleton element a of real numbers, this set is empty.

Because there is no increment so if $a\in \mathbb R$ then the set (a,a] represents singleton sets, and singleton sets are empty so is (a,a].

(iv) (a,a)

Means given set excluding a by left as well as right.

Since there is a singleton element a of real numbers, this set is empty.

Its elements are of the form.

$( 1, 1),(-1,-1],(2,2],(-2,-2],(3,3],(-3,-3],........$

Because there is no increment so if $a\in \mathbb R$ then the set (a,a) represents singleton sets, and singleton sets are empty, so is (a,a).

(v) (a,b) where a>b.

Which indicate the interval containing a, b such that increment of x is always greater than increment of y which not take x and y by any side of the interval.

That is the graph is bounded by value of a and it contains elements like it we fixed a=5 then,

$(a,b)=\{(5,0),(5,1),(5,2).....\}$ e.t.c

So this set is connected and we know singletons are connected in $\mathbb R$ . Hence given set is empty.

(vi) [a,b] where $a\leq b$ .

Which indicate the interval containing a, b such that increment of x is always greater than increment of y which include both x and y.

That is the graph is bounded by value of a and it contains elements like it we fixed a=5 then,

$[a,b]=\{[5,0],[5,1],[5,2].....\}$ e.t.c

So this set is connected and we know singletons are connected in $\mathbb R$ . Hence given set is empty.

8 0

2 years ago