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2 years ago

-4(5x)=12 is it true false or open?

Mathematics

2 answers:

Ipatiy [6.2K]2 years ago

8 0

False if that's what you're asking

icang [17]2 years ago

7 0

Answer:

True when x = -3/5; false otherwise

Step-by-step explanation:

Note that -4(5x) = 12 => -20x = 12, so that x = -3/5. The given equation is solveable, and the solution is -3/5. Asking whether -4(5x)=12 is "true" or "false" is pointless; this equation is conditionally true, with that truth hinging on x = -3/5. Similarly, -4(5x)=12 is not describable by "open."

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15 points Help asap thanks so much

lyudmila [28]

The answer is inconsistent

5 0

2 years ago

A dairy wishes to mix together 1,000 pints of milk that contains 8% butterfat.If the mixture is to be made from milk containing

3241004551 [841]

Answer:

800 pints of 5% butterfat and 200 pints of 20% butterfat is needed.

Step-by-step explanation:

Given:

A dairy wishes to mix together 1,000 pints of milk that contains 8% butterfat.

If the mixture is to be made from milk containing 5% butterfat, and cream

containing 20% butterfat.

Now, to find the pints of each is needed.

Let the pints of 5% butterfat be $x.$

And the pints of 20% butterfat be $y.$

So, the total pints of milk:

$x+y=1000$

$x=1000-y$ ......(1)

As, given:

1,000 pints of milk that contains 8% butterfat.

The mixture of milk containing 5% butterfat, and cream containing 20% butterfat.

According to question:

$5(x)+20(y)=8(1000)$

$5x+20y=8000$

Substituting the value of $x$ from equation (1):

$5(1000-y)+20y=8000$

$5000-5y+20y=8000$

$5000+15y=8000$

Subtracting both sides by 5000 we get:

$15y=3000$

Dividing both sides by 15 we get:

$y=200.$

The pints of 20% butterfat = 200 pints.

Now, to get the pints of 5% butterfat we put the value of $y$ in equation (1):

$x=1000-y$

$x=1000-200$

$x=800.$

The pints of 5% butterfat = 800 pints.

Therefore, 800 pints of 5% butterfat and 200 pints of 20% butterfat is needed.

8 0

2 years ago

Solve for x and y:

Pachacha [2.7K]

Answer:

A= I think because it is definite not c so there is a solution differently I know so that is my final answer.

Sorry if I am wrong.

4 0

3 years ago

16y5+4y5-12y3+15y3? What is the simplified form?

Eddi Din [679]

Answer:

$20y^5+3y^3$

Step-by-step explanation:

$16y^5 + 4y^5 -12y^3 + 15y^3$

All we can do here is combine like terms (add those that have the same the same exponent)

$20y^5+3y^3$

Good luck!

3 0

2 years ago

Read 2 more answers

Remember to show work and explain. Use the math font.

vichka [17]

Answer:

$\large\boxed{1.\ f^{-1}(x)=\sqrt[12]{3^x}}\\\\\boxed{2.\ f^{-1}(x)=\sqrt[4]{3^x}}\\\\\ \boxed{3.\ f^{-1}(x)=\sqrt[3]{4^{7-x}}}$

Step-by-step explanation:

$(a^n)^m=a^{nm}\\\\\log_ab=c\iff a^c=b\\\\a^{\log_ax}=x\\\\n\log_ab=\log_ab^n\\\\\log_ab+\log_ac=\log_a(bc)\\============================\\\\1.\\y=3\log_3x^4\to y=\log_3(x^4)^3\to y=\log_3x^{12}$

$2.\\y=\log_3x^4\\\\\text{Exchange x and y. Solve for y:}\\\\\log_3y^4=x\Rightarrow3^{\log_3y^4}=3^x\Rightarrow y^{4}=3^x\\\\y=\sqrt[4]{3^x}\\-------------------------$

$3.\\y=-\log_4x^3+7\\\\\text{Exchange x and y. Solve for y:}\\\\-\log_4y^3+7=x\qquad\text{subtract 7 from both sides}\\\\-\log_4 y^3=x-7\qquad\text{change the signs}\\\\\log_4y^3=7-x\Rightarrow4^{\log_4y^3}=4^{7-x}\\\\y^3=4^{7-x}\Rightarrow y=\sqrt[3]{4^{7-x}}$

7 0

2 years ago