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3 years ago

519 divided by 6 and 915 divided by 7 and 439 divided by 7 and 812 divided by 9 which one does not have a two digit quotient.

Mathematics

1 answer:

Setler [38]3 years ago

5 0

519/6=86.5

915/7=130.7142

439/7=62.7142

812/9=90.2222

so i think its the 1st on but i may be wrong.

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6. Which expression is equivalent to -21 + 28x?

Rina8888 [55]

Solution,
-28x^2+35x
take the common,
-7x(4x-5)
The answer is -7x(4x-5)
Here,
If we solve,
-7x(4x-5)
Then, the answer would be:
= -7x*4x+7x*5
= -28x^2+35x
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3 0

3 years ago

Use the function f(x) to answer the questions.

frutty [35]

Part A

$-16x^2 + 60x+16=0\\\\4x^2 - 15x-4=0\\\\(4x+1)(x-4)=0\\\\x=-\frac{1}{4}, 4$

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The vertex of the graph will be a maximum because the leading coefficient is negative.

The x coordinate of the vertex is $-\frac{60}{2(-16)}=\frac{15}{8}$ .

When x=15/8, $f\left(\frac{15}{8} \right)=\frac{289}{4}$ .

So, the vertex has coordinates $\left(\frac{15}{8}, \frac{289}{4} \right)$

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The graph is in the attached image.

3 0

2 years ago

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Sever21 [200]

Answer:

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3 years ago

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

Maru [420]

Parameterize $S{/tex] by[tex]\vec s(u,v)=u\,\vec\imath+v\,\vec\jmath+(8-u^2-v^2)\,\vec k$

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Take the normal vector to $S$ to be

$\vec s_u\times\vec s_v=2u\,\vec\imath+2v\,\vec\jmath+\vec k$

Then the flux of $\vec F$ across $S$ is

$\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\int_0^1\int_0^1\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^1\int_0^1(uv\,\vec\imath+v(8-u^2-v^2)\,\vec\jmath+u(8-u^2-v^2)\,\vec k)\cdot(2u\,\vec\imath+2v\,\vec\jmath+\vec k)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^1\int_0^1\bigg(2u^2v+(u+2v^2)(8-u^2-v^2)\bigg)\,\mathrm du\,\mathrm dv=\boxed{\frac{1553}{180}}$

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3 years ago

URGENT

AlekseyPX

Answer:

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2 years ago