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2 years ago

Four watermelons cost $22. Complete the table.

Mathematics

1 answer:

Elodia [21]2 years ago

6 0

Answer:

which table xD

Step-by-step explanation:

:ooo

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Jacques and Aaron both started this year at 5.65 feet tall. Jacques has grown 0.2 feet since then. Aaron has grown twice that am

OLga [1]

It is 5.69 positively sure of it good luck on youre quiz if youre taking one!

7 0

2 years ago

-f(3f-7)=0 solve equation

kenny6666 [7]

Answer:

F=0 or (7/3)

Step-by-step explanation:

When F is 0, the equation reads -0(3(0)-7)=0. The outside 0 will multiply by everything and make it equal 0. When F is 7/3, the inside of the parenthesis read 3(7/3)-7. This equals 7-7. It'll end up being (7/3)0, which equals 0.

3 0

2 years ago

What is the result of substituting for y in the bottom equation y=x+3 y=x^2+2x-4

8_murik_8 [283]

The solutions are (2.1925, 5.1925) and (-3.1925, -0.1925)

Solution:

Given that,

$y = x + 3 ------- eqn 1\\\\y = x^2 + 2x - 4 ----- eqn 2$

We have to substitute eqn 1 in eqn 2

$x + 3 = x^2 + 2x - 4$

$\mathrm{Switch\:sides}\\\\x^2+2x-4=x+3\\\\\mathrm{Subtract\:}3\mathrm{\:from\:both\:sides}\\\\x^2+2x-4-3=x+3-3\\\\\mathrm{Simplify}\\\\x^2+2x-7=x\\\\\mathrm{Subtract\:}x\mathrm{\:from\:both\:sides}\\\\x^2+2x-7-x=x-x\\\\\mathrm{Simplify}\\\\x^2+x-7=0$

$\mathrm{Solve\:with\:the\:quadratic\:formula}\\\\\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\mathrm{For\:}\quad a=1,\:b=1,\:c=-7\\\\x =\frac{-1\pm \sqrt{1^2-4\cdot \:1\left(-7\right)}}{2\cdot \:1}$

$x = \frac{-1 \pm \sqrt{ 1 + 28}}{2}\\\\x = \frac{ -1 \pm \sqrt{29}}{2}$

$x = \frac{ -1 \pm 5.385 }{2}\\\\We\ have\ two\ solutions\\\\x = \frac{ -1 + 5.385 }{2}\\\\x = 2.1925$

$Also\\\\x = \frac{ -1 - 5.385 }{2}\\\\x = -3.1925$

Substitute x = 2.1925 in eqn 1

y = 2.1925 + 3

y = 5.1925

Substitute x = -3.1925 in eqn 1

y = -3.1925 + 3

y = -0.1925

Thus the solutions are (2.1925, 5.1925) and (-3.1925, -0.1925)

6 0

3 years ago

1. Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial i

Katena32 [7]

Polynomial with real coefficients always has even number of complex roots. We know that one of them is 2 + 5i so the second one will be 2 - 5i and:

$f(x)=\big(x-4\big)\big(x-(-8)\big)\big(x-(2+5i)\big)\big(x-(2-5i)\big)=\\\\=(x-4)(x+8)(x-2-5i)(x-2+5i)=\\\\=(x^2-4x+8x-32)(x^2-2x+5ix-2x+4-10i-5ix+10i-25i^2)\\\\=\big(x^2+4x-32\big)\big(x^2-4x+4-25\cdot(-1)\big)=\\\\=(x^2+4x-32)(x^2-4x+29)=\\\\=x^4-4x^3+29x^2+4x^3-16x^2+116x-32x^2+128x-928=\\\\=\boxed{x^4-19x^2+244x-928}$

Answer B.

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2 years ago

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seropon [69]

Answer:

the answer is D. 5.6

Step-by-step explanation:

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2 years ago

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