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4 years ago

12

I really need the answer I’ve been stuck on this all day

2 answers:

zimovet [89]4 years ago

7 0

Hello!

You do 72/6 which is 12.

So the answer is 12

Hope this helps!

Nikitich [7]4 years ago

7 0

Formula A=bh
The area is 72
72=6*b
Base= 12

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19. Mr. Green teaches band, choir, and math. This year, he has 57 students that take at least one of (3 points)

kvasek [131]

Answer:

The answer is "17 students with Mr. Green take exactly two courses".

Step-by-step explanation:

Following are the calculation to this question:

Alone band 11

Alone chorus 17

Choir and band alone 4

7 Maths alone and the band

Choir and Math alone 6

Chorus and Math Band 3

mathematics alone 9 +11+4+10+6+17 + M = 57

so

2 classes = 4+7+6 = 17

So, there are 17 students with Mr. Green take exactly two courses.

8 0

3 years ago

Which expressions are equivalent to 3g+6(-g+(-5))3g+6(−g+(−5))3, g, plus, 6, left parenthesis, minus, g, plus, left parenthesis,

liberstina [14]

Answer: none of the above

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

The function, f(x) = –2x2 + x + 5, is in standard form. The quadratic equation is 0 = –2x2 + x + 5, where a = –2, b = 1, and c =

maw [93]

Answer:

To solve for the zeros of the function equate f(x) = 0

That's

- 2x² + x + 5 = 0

Using the quadratic formula

$x = \frac{ - b± \sqrt{ {b}^{2} - 4ac} }{2a}$

a = - 2 b = 1 c = 5

And from the question

b² - 4ac = 41

So we have

$x = \frac{ - 1± \sqrt{41} }{2( - 2)} = \frac{ - 1± \sqrt{41} }{ - 4}$

$x = \frac{1± \sqrt{41} }{4}$

We have the final answer as

$x = \frac{1 + \sqrt{41} }{4} \: \: \: \: or \: \: \: \: x = \frac{1 - \sqrt{41} }{4}$

Hope this helps you

4 0

3 years ago

Read 2 more answers

7x + 2y = 17<br> (3x +2y = 3)

Gnoma [55]

Step-by-step explanation:

Multiply second equation by -1

-1 × (3x + 2y) = 3

-3x -3y = -3 now add this to the first equation

7x +2y = 17

(-3x-2y)= -3

7x +2y -3x -2y = 17 -3 (2y will eliminate -2y)

4x = 14

x = 7/2 now by using this information we found we can calculate the value for y

if 3x + 2y = 3 (replace x by 7/2)

3 ×(7/2)+2y = 3 ➡ 21/3 + 2y = 3 ➡ 2y = 3 - 21/3 ➡ 2ty = 15/2 ➡ y = 15/4

8 0

3 years ago

Ejercicio 3. Aplica las leyes de los exponentes de potencia elevada a otra potencia y de producto,

OLga [1]

Los resultados son los siguientes:

$2097152$
$\frac{1}{729}$
$b^{12}$
$x^{24}$
$x^{\frac{6}{5} }$
$3600$
$a^{6}\cdot b^{10}\cdot c^{-8}$
$200\cdot x^{19}$
$108\cdot y^{6}$

En este ejercicio debemos aplicar cualesquiera de las siguientes propiedades asociadas a los exponentes:

(i) $(a^{b})^{c} = a^{b\cdot c}$

(ii) $a^{c}\cdot b^{c} = (a\cdot b)^{c}$

(iii) $a^{-b} = \frac{1}{a^{b}}$

(iv) $a^{b} = a^{b}$ , donde $a$ es negativo o positivo y $b$ es par.

(v) $(-a)^{b} = -a^{b}$ , donde $a$ es positivo y $b$ es impar.

(vi) $a^{b}\cdot a^{c} = a^{b+c}$

A continuación, tenemos los siguientes resultados:

a) $(2^{3})^{7} = 8^{7} = 2097152$

b) $(3^{3})^{-2} = 3^{-6} = \frac{1}{729}$

c) $(b^{3})^{4}=b^{12}$

d) $[(x^{2})^{3}]^{4} = (x^{6})^{4} = x^{24}$

e) $\left(x^{\frac{3}{5} }\right)^{2} = x^{\frac{6}{5} }$

f) $(3\cdot 4\cdot 2)^{5} = 3^{2}\cdot 4^{2}\cdot 5^{2} = 9\cdot 16\cdot 25 = 3600$

g) $(a^{3}\cdot b^{5}\cdot c^{-4})^{2} = (a^{3})^{2}\cdot (b^{5})^{2}\cdot (c^{-4})^{2} = a^{6}\cdot b^{10}\cdot c^{-8}$

h) $(2\cdot x^{5})^{3}\cdot (-5\cdot x^{2})^{2} = (2^{3}\cdot x^{15})\cdot [(-5)^{2}\cdot x^{4}] = (8\cdot x^{15})\cdot (25\cdot x^{4}) = (8\cdot 25)\cdot x^{19} = 200\cdot x^{19}$

i) $\left(3\cdot y^{\frac{2}{3} }\right)^{3}\cdot (2\cdot y^{2})^{2} = (27\cdot y^{2})\cdot (4\cdot y^{4}) = 108\cdot y^{6}$

Invitamos cordialmente a consultar esta pregunta sobre potencias: brainly.com/question/24239709

8 0

3 years ago