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2 years ago

Write the word sentence as an inequality.

Mathematics

2 answers:

Nataly_w [17]2 years ago

8 0

Answer:

2.3 + w > 18

Step-by-step explanation:

w is added to 2.3 is said first so it's put first. This is put in front of the symbol. After the symbol is 18.

So far the equation looks like:

2.3 + w _ 18

The symbol is said as more than. The left side is more than the right.

This symbol means more than: >

There is no line under it because the line means "or equal to"

Final answer: 2.3 + w > 18

Mashutka [201]2 years ago

6 0

2.3+w>18. This is most likely the correct answer

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Which steps are correct? Check all that apply.

sladkih [1.3K]

Answer:

Substitute the known values into the formula: 140 = 7(w)(5).

Solve for the unknown: w = 4 m.

Check:

140 = 7(w)(5) => w = 140/(5 x 7) = 140/35 = 4. => Correct.

Substitute the known values into the formula:

7 = 140(w)(5).

Solve for the unknown: w = 7 m.

Check:

7 = 140(w)(5) => w = 7/(140 x 5) = 7/700 = 1/100. => Incorrect.

Substitute the known values into the formula:

5 = 7(w)(140).

Solve for the unknown: w = 2 m.

Check: 5 = 7(w)(140) => w = 5/(7 x 140) = 5/840 = 1/168. => Incorrect.

Hope this helps!

7 0

2 years ago

Plssss helppppp and thanks you if answers are actually true then I give brain:)

san4es73 [151]

The 3rd one and the last one.

7 0

2 years ago

F(x) = x2 – 3x – 2 is shifted 4 units left. The result is g(x). What is g(x)?

Ray Of Light [21]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\

\begin{array}{rllll} 
% left side templates
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}
\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\end{array}$

$\bf \begin{array}{llll}


\bullet \textit{ vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\\\
\qquad if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}
\end{array}$

now... your expression is not in vertex form, that's ok... to do a horizontal shift to the left by 4 units, we can simply, add the C and B component to the "x" variable, C=4, B =1, that way the horizontal shift of C/B or 4/1 is just +4, giving us a horizontal shift to the left

$\bf f(x)=x^2-3x-2\impliedby \textit{let's change that for }f(1x+4)
\\\\\\
f(1x+4)=(1x+4)^2-3(1x+4)-2
\\\\\\
f(x+4)=x^2+8x+16-3x-12-2
\\\\\\
f(x+4)=x^2-5x+2\impliedby g(x)$

8 0

2 years ago

Read 2 more answers

HARD QUADRATICS!!!! A quadratic of the form $-2x^2 + bx + c$ has roots of $x = 3 + \sqrt{5}$ and $x = 3 - \sqrt{5}.$ The graph o

iren2701 [21]

Answer:

hold up this will take a min to work out

Step-by-step explanation:

5 0

2 years ago

The spherical balloon is inflated at the rate of 10 m³/sec. Find the rate at which the surface area is increasing when the radiu

rjkz [21]

The balloon has a volume $V$ dependent on its radius $r$ :

$V(r)=\dfrac43\pi r^3$

Differentiating with respect to time $t$ gives

$\dfrac{\mathrm dV}{\mathrm dt}=4\pi r^2\dfrac{\mathrm dr}{\mathrm dt}$

If the volume is increasing at a rate of 10 cubic m/s, then at the moment the radius is 3 m, it is increasing at a rate of

$10\dfrac{\mathrm m^3}{\mathrm s}=4\pi (3\,\mathrm m)^2\dfrac{\mathrm dr}{\mathrm dt}\implies\dfrac{\mathrm dr}{\mathrm dt}=\dfrac5{18\pi}\dfrac{\rm m}{\rm s}$

The surface area of the balloon is

$S(r)=4\pi r^2$

and differentiating gives

$\dfrac{\mathrm dS}{\mathrm dt}=8\pi r\dfrac{\mathrm dr}{\mathrm dt}$

so that at the moment the radius is 3 m, its area is increasing at a rate of

$\dfrac{\mathrm dS}{\mathrm dt}=8\pi(3\,\mathrm m)\left(\dfrac5{18\pi}\dfrac{\rm m}{\rm s}\right)=\dfrac{20}3\dfrac{\mathrm m^2}{\rm s}$

4 0

2 years ago