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3 years ago

How much more than 1/4 cup is 3/4 a cup? Write in simplest form.

Mathematics

2 answers:

GenaCL600 [577]3 years ago

8 0

3/4 cup is 2/4 cup greater than 1/4 cup. In simplest terms 2/4 cup would be written as 1/2 cup.

Eddi Din [679]3 years ago

5 0

3/4 of a cup is 2/4, or 1/2 of a cup more than 1/4 of a cup.

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I need help with question number 1 and question number 2 please

Ivenika [448]

1. It is a quadratic equation.

Explanation:

A quadratic equation is any equation that can be rearranged in standard form (ax^2 + bx + c). That equation can be rewritten as -x^2 + 1, where a is -1, b is 0, and c is 1.

2. 4x^2, 3x, and 5.

Explanation:

A quadratic term is ax^2 in the formula ax^2 + bx + c. A linear term is bx in the formula, and a constant is c in the formula.

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3 years ago

A) (+4) + ( - 3) b) (-2) + (-3) c) (+2) + (+4) d) (-6) + (+2)

dolphi86 [110]

Answer:

c ) 6

d) 1 0 this are correct answer ok

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3 years ago

Negative six and one third plus twenty over three

nydimaria [60]

Answer:

1/3

Step-by-step explanation:

-6+(-1/3)= -19/3

-19/3 + 20/3 = 1/3

6 0

4 years ago

-5(-4m - 6)

Natalka [10]

Answer:

1 is the correct answer

Step-by-step explanation:

4 0

3 years ago

What is 2 (log Subscript 3 Baseline 8 + log Subscript 3 Baseline z) minus log Subscript 3 Baseline (3 Superscript 4 Baseline min

Ilya [14]

Answer:

Value of expression in single logarithm is $\log_3\left(2z^2\right)$ .

Step-by-step explanation:

Given expression is,

$2\left(\log_3\left(8\right)+\log_3\left(z\right)\right)-\log_3\left(3^4-7^2\right)$

Now using logarithmic rule to solve the expression as follows,

Applying product rule of logarithmic,

$\log_c\left(a\right)+\log_c\left(b\right)=\log_c\left(ab\right)$

Therefore,

$2\log_3\left(8z\right)-\log_3\left(3^4-7^2\right)$

Applying power rule of logarithmic,

$a\log_c\left(b\right)=\log_c\left(b^a\right)$

Therefore,

$\log_3\left(\left(8z\right)^2\right)-\log_3\left(3^4-7^2\right)$

$\log_3\left(\left(64z^2\right)\right)-\log_3\left(3^4-7^2\right)$

Applying quotient rule of logarithmic,

$\log_c\left(a\right)-\log_c\left(b\right)=\log_c\left(\frac{a}{b}\right)$

Therefore,

$\log_3\left(\dfrac{\left(64z^2\right)^2}{3^4-7^2}\right)$

Simplifying,

$\log_3\left(\dfrac{\left(64z^2\right)^2}{81-49}\right)$

$\log_3\left(\dfrac{\left(64z^2\right)^2}{32}\right)$

$\log_3\left(2z^2\right)$

Therefore value of expression is $\log_3\left(2z^2\right)$

6 0

4 years ago

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