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elena-s [515]
3 years ago
13

Carlos is using a number line to add another integer to –4. He begins by showing –4 on the number line, as shown below. Carlos a

dds the other integer on the number line and says that the sum is negative.Which is true of the number Carlos added to –4?
Mathematics
2 answers:
Marrrta [24]3 years ago
7 0

Answer:

The number could be 3 or any integer less than 3.

Step-by-step explanation:

i took the test

kifflom [539]3 years ago
5 0
The integer is 3 or less than 3.
You might be interested in
PLEASE LOOK AT PHOTO! WHOEVER ANSWERS CORRECTLY I WILL MARK BRAINIEST!!
antoniya [11.8K]

Answer:

33 degree

Step-by-step explanation:

Let m1 be x

m2= x+1

A/Q

--> m1+m2= 67

--> x+x+1= 67

--> 2x+1=67

--> 2x= 67-1

--> 2x=66

--> x= 66/2

--> x= 33

So, m1= 33 degree

PLEASE MARK ME AS BRAINLIEST:)

7 0
3 years ago
What is the mode of this data set: 17, 13, 18, 20, 17, 15, 12<br><br> A. 19<br><br> B. 17
telo118 [61]
<h3>Answer: B. 17</h3>

The mode is the most frequent or most occurring value. The value 17 shows up the most times, at two times, so this is the mode. The value 19 is not even part of the given set of values, so you can immediately cross of choice A.

Side note: it is possible for a set to have more than one mode, and it's also possible to not have a mode at all.

7 0
4 years ago
State if the two triangles are congruent. If they are, state how you know.
Margarita [4]

Answer: yes, aas

Step-by-step explanation:

They have the two angles, and then the side after the angles

7 0
3 years ago
To show me similarity to this statement, how can it be done?
Alenkasestr [34]

We start with the expression at the left of the equation.

We can combine the terms as:

\begin{gathered} \frac{2+\sqrt[]{3}}{\sqrt[]{2}+\sqrt[]{2-\sqrt[]{3}}}-\frac{2-\sqrt[]{3}}{\sqrt[]{2}-\sqrt[]{2-\sqrt[]{3}}} \\ \frac{2+\sqrt[]{3}}{\sqrt[]{2}+\sqrt[]{2-\sqrt[]{3}}}\cdot\frac{(\sqrt[]{2}-\sqrt[]{2-\sqrt[]{3}})}{(\sqrt[]{2}-\sqrt[]{2-\sqrt[]{3}})}-\frac{2-\sqrt[]{3}}{\sqrt[]{2}-\sqrt[]{2-\sqrt[]{3}}}\cdot\frac{(\sqrt[]{2}+\sqrt[]{2-\sqrt[]{3}})}{(\sqrt[]{2}+\sqrt[]{2-\sqrt[]{3}})} \\ \frac{(2+\sqrt[]{3})\cdot(\sqrt[]{2}-\sqrt[]{2-\sqrt[]{3}})-(2-\sqrt[]{3})\cdot(\sqrt[]{2}+\sqrt[]{2-\sqrt[]{3}})}{(\sqrt[]{2}+\sqrt[]{2-\sqrt[]{3}})(\sqrt[]{2}-\sqrt[]{2-\sqrt[]{3}})} \end{gathered}

We can now apply the distributive property for the both the numerator and denominator. We can see also that the denominator is the expansion of the difference of squares:

\begin{gathered} \frac{(2+\sqrt[]{3})\cdot(\sqrt[]{2}-\sqrt[]{2-\sqrt[]{3}})-(2-\sqrt[]{3})\cdot(\sqrt[]{2}+\sqrt[]{2-\sqrt[]{3}})}{(\sqrt[]{2})^2-(\sqrt[]{2-\sqrt[]{3}}))^2} \\ \frac{(2+\sqrt[]{3})\cdot(\sqrt[]{2}-\sqrt[]{2-\sqrt[]{3}})+(\sqrt[]{3}-2)\cdot(\sqrt[]{2}+\sqrt[]{2-\sqrt[]{3}})}{2^{}-(2-\sqrt[]{3})^{}} \\ \frac{\sqrt[]{2}\cdot(2+\sqrt[]{3})-\sqrt[]{2-\sqrt[]{3}}\cdot(2+\sqrt[]{3})+\sqrt[]{2}\cdot(\sqrt[]{3}-2)+\sqrt[]{2-\sqrt[]{3}}\cdot(\sqrt[]{3}-2)}{2-2+\sqrt[]{3}} \\ \frac{\sqrt[]{2}(2+\sqrt[]{3}+\sqrt[]{3}-2)+\sqrt[]{2-\sqrt[]{3}}(-2-\sqrt[]{3}+\sqrt[]{3}-2)}{\sqrt[]{3}} \\ \frac{\sqrt[]{2}(2\sqrt[]{3})+\sqrt[]{2-\sqrt[]{3}}(-4)}{\sqrt[]{3}} \\ 2\sqrt[]{2}-4\frac{\sqrt[]{2-\sqrt[]{3}}}{\sqrt[]{3}} \end{gathered}

We then can continue rearranging this as:

7 0
1 year ago
Sarai is mixing a solution. She pours all the liquid from a full small beaker into a larger beaker. The liquid fills the large b
daser333 [38]

Answer:

2000 ml

Step-by-step explanation:

Given: Sarai pours all the liquid from a full small beaker into a  larger beaker.

           The liquid fills the large beaker to 15% of its capacity.

            The small beaker holds 300 ml.

Lets assume capacity of large beaker to hold be "x".

As given, Sarai pours all the liquid from a full small beaker into a larger beaker

∴ 15\% \times x= 300\ ml

⇒ 0.15x= 300

Dividing both side by 0.15

⇒x= \frac{300}{0.15}

∴ x= 2000\ ml

Hence, the large beaker can hold 2000 ml of liquid.

8 0
3 years ago
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