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3 years ago

Solve this please!!!!!

Mathematics

2 answers:

Leya [2.2K]3 years ago

6 0

Answer:

answer 5,7

Step-by-step explanation:

algol [13]3 years ago

4 0

Omg we literally learn the same thing ahahah I’m in precal 11

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1/3n-7/12=-3/8 n=5/72 n=-5/72 n=5/8 n=-2 7/8

prisoha [69]

Given the equation;

$\frac{1}{3}n-\frac{7}{12}=-\frac{3}{8}$

To solve, let us multiply through by the least common multiple of the the denominator of the three fractions wich is 24;

3 0

1 year ago

Find the GCF of the list below.<br> - 20x^3y. 4x^4 y^2. 60x^4y

iogann1982 [59]

Answer:

Check photo

Step-by-step explanation:

8 0

3 years ago

Which value of the 9 ten times in the number 920

umka21 [38]

Answer:

9000. the value of 9 in 920 is 900. and ten times anything is basically just adding a 0. so ten times 900 is 9000.

Step-by-step explanation:

Please mark brainliest and have a great day!

3 0

3 years ago

What is this in simplest form?

Norma-Jean [14]

Answer:

1/4

Step-by-step explanation

you divide 25 divided by 25 the also divide 100 divided by 25 and u get 1/4

7 0

3 years ago

Read 2 more answers

2x² - 5x+1 has roots alpha and beta. Find alpha⁴+beta⁴ without solving the equation.

shepuryov [24]

Answer:

Step-by-step explanation:

$\alpha+\beta=\dfrac{5}{2} \\\\\alpha*\beta=\dfrac{1}{2} \\\\\alpha^2+\beta^2=\dfrac{21}{4} \ (see\ previous\ post)\\\\(\alpha+\beta)^4=\dfrac{625}{16} \\\\=\alpha^4+\beta^4+4*\alpha^3*\beta+6*\alpha^2*\beta^2+4*\alpha*\beta^3\\\\=\alpha^4+\beta^4+4*(\alpha*\beta)(\alpha^2+\beta^2)+6*\alpha^2*\beta^2\\\\\alpha^4+\beta^4=(\alpha+\beta)^4-4*(\alpha*\beta)(\alpha^2+\beta^2)-6*\alpha^2*\beta^2\\\\= \dfrac{625}{16} -4*\dfrac{1}{2} *\dfrac{21}{4} -6*(\dfrac{1}{2})^2 \\\\$

$= \dfrac{625}{16}- \dfrac{168}{16}-\dfrac{24}{16}\\\\\\= \dfrac{433}{16}$

7 0

3 years ago