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Math Problems
Grade 6
Inequalities with addition and subtraction of like and unlike fractions
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Digital SAT Math
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COURSE: DIGITAL SAT MATH > UNIT
4
4
4
Quiz
2
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2
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Quiz
2
2
2
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Foundations: Advanced math
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Which of the following ordered pairs are solutions to the system graphed?
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Test prep
>
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>
Digital SAT Math
>
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>
Foundations:
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Choose
1
1
1
answer: Advanced math > Quiz
2
2
2
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(A)
(
3
,
0
)
(3,0)
(
3
,
0
)
and
(
−
2
,
0
)
(-2,0)
(
−
2
,
0
)
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(c)
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Get tutor help
Which sign makes the statement true?
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6
8
−
2
8
\frac{6}{8} - \frac{2}{8}
8
6
−
8
2
?
\text{?}
?
1
8
+
3
8
\frac{1}{8} + \frac{3}{8}
8
1
+
8
3
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Choices:
\newline
(A)
>
>
>
\newline
(B)
<
<
<
\newline
(C)
=
=
=
Get tutor help
Which sign makes the statement true?
\newline
9
11
−
3
11
\frac{9}{11} - \frac{3}{11}
11
9
−
11
3
?
\text{?}
?
9
11
−
2
11
\frac{9}{11} - \frac{2}{11}
11
9
−
11
2
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Choices:
\newline
(A)
>
>
>
\newline
(B)
<
<
<
\newline
(C)
=
=
=
Get tutor help
Which sign makes the statement true?
\newline
6
12
+
3
12
\frac{6}{12} + \frac{3}{12}
12
6
+
12
3
?
\text{?}
?
7
12
+
4
12
\frac{7}{12} + \frac{4}{12}
12
7
+
12
4
\newline
Choices:
\newline
(A)
>
>
>
\newline
(B)
<
<
<
\newline
(C)
=
=
=
Get tutor help
Which sign makes the statement true?
\newline
5
9
+
1
9
\frac{5}{9} + \frac{1}{9}
9
5
+
9
1
?
6
9
+
2
9
\frac{6}{9} + \frac{2}{9}
9
6
+
9
2
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Choices:
\newline
(A) >
\newline
(B) <
\newline
(C) =
\newline
Get tutor help
Which sign makes the statement true?
\newline
6
9
−
3
9
\frac{6}{9} - \frac{3}{9}
9
6
−
9
3
?
\text{?}
?
1
9
+
3
9
\frac{1}{9} + \frac{3}{9}
9
1
+
9
3
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Choices:
\newline
(A)
>
>
>
\newline
(B)
<
<
<
\newline
(C)
=
=
=
Get tutor help
Which sign makes the statement true?
\newline
2
5
+
1
5
\frac{2}{5} + \frac{1}{5}
5
2
+
5
1
?
1
5
+
2
5
\frac{1}{5} + \frac{2}{5}
5
1
+
5
2
\newline
Choices:
\newline
(A)
>
>
>
\newline
(B)
<
<
<
\newline
(C)
=
=
=
Get tutor help
Which sign makes the statement true?
\newline
9
12
−
3
12
\frac{9}{12} - \frac{3}{12}
12
9
−
12
3
?
10
12
−
2
12
\frac{10}{12} - \frac{2}{12}
12
10
−
12
2
\newline
Choices:
\newline
(A)
>
>
>
\newline
(B)
<
<
<
\newline
(C)
=
=
=
Get tutor help
