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2 years ago

Please help me with this question

Mathematics

1 answer:

Alenkasestr [34]2 years ago

3 0

Answer:

$\sqrt{8 + y}$

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PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!<br><br> What is the value of x?

MA_775_DIABLO [31]

Answer: $\bold{C)\ -\dfrac{1}{2}}$

<u>Step-by-step explanation:</u>

$.\qquad \dfrac{x+2}{15}=\dfrac{x+1}{5}\\\\\\\rightarrow (15)\dfrac{x+2}{15}=\dfrac{x+1}{5}(15)\\\\\\\rightarrow x+2=3(x+1)\\\\\rightarrow x+2=3x+3\\\\\rightarrow \qquad 2=2x+3\\\\\rightarrow \qquad -1=2x\\\\\rightarrow \qquad -\dfrac{1}{2}=x$

8 0

3 years ago

keiko has bought 36pounds of dog food, she feed the dog 3/4 pound of food each meal. how many meals would this last

exis [7]

Answer:

48 meals

Step-by-step explanation:

36lbs ÷ 3/4lbs = 48

5 0

3 years ago

What is y=2x+4 and 3x+y=9 substitution method

soldier1979 [14.2K]

You plug in 2x + 4 into the y for the other equation (3x +y ) and solve for it

8 0

3 years ago

Which line has an equation of y=-5x+4 in slope-intercept form?

stellarik [79]

Answer:

Last choice.

Step-by-step explanation:

To find the option that is equivalent to this equation, we can begin by narrowing down our options. The slope contains a negative number, meaning the second point must have a lower y value than the first.

Using this criteria, we can eliminate the first and the third answer. Now, we can test the second and fourth answer.

We can test using the slope formula and plugging in values. When the slope formula: $m= \frac{y_{2}-y_{1} }{x_{2} -x_{1} }$

We get a slope of -5 for both of these equations. Let's plug in the point values to check whether they work:

Choice 2:

-24 = -5(4) + 4

-24 = -20 + 4

-24 ≠ -16

Choice 4:

-16 = -5(4) + 4

-16 = -20 + 4

-16 = -16

Therefore, choice 4 is correct!

4 0

2 years ago

F(x)=2x^2-3x+7 evaluate f(-2)

Vilka [71]

Answer:

Step-by-step explanation:

Find two linear functions p(x) and q(x) such that (p (f(q(x)))) (x) = x^2 for any x is a member of R?

Let p(x)=kpx+dp and q(x)=kqx+dq than

f(q(x))=−2(kqx+dq)2+3(kqx+dq)−7=−2(kqx)2−4kqx−2d2q+3kqx+3dq−7=−2(kqx)2−kqx−2d2q+3dq−7

p(f(q(x))=−2kp(kqx)2−kpkqx−2kpd2p+3kpdq−7

(p(f(q(x)))(x)=−2kpk2qx3−kpkqx2−x(2kpd2p−3kpdq+7)

So you want:

−2kpk2q=0

and

kpkq=−1

and

2kpd2p−3kpdq+7=0

Now I amfraid this doesn’t work as −2kpk2q=0 that either kp or kq is zero but than their product can’t be anything but 0 not −1 .

Answer: there are no such linear functions.

6 0

2 years ago