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3 years ago

Some one please help me with this question please

Mathematics

1 answer:

melamori03 [73]3 years ago

4 0

4.5 miles per hour
Yasmine's speed is faster than Jermaine's because when you subtract the two numbers, you get 4.5 miles per hour.

I hope I helped

Please make me brainliest

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Joel wanted his mom to buy a super-sized box of his favorite cereal, but his mom didn't think it would fit in her cupboard. The

Lorico [155]

<h3><u>Given</u>:-</h3>

Volume 6,900 cm^3
Length = 23 cm
Width = 10cm

Height of box ??

<h3><u>Solution</u>:-</h3>

$\\\diamond{\underline{\underline{\sf {\;\; Calculating\; Height\; of \; box :-}}}}\\$

$\\ \pink \star \: \: {\underline{\boxed {\bf{ Volume_{ \red{ \{Rectangle \}}} \: = l \times w \times h }}} }\; \red\bigstar \\$

Where,

» l denotes Length
» w denotes Width
» h denotes Height

$\\ \sf \implies \: {Volume_{ \red{ \{Rectangle \}}} \: = l \times w \times h} \\$

$\\ \sf \implies \: { 6,900 \: = \: 23 \: \times \: 10 \: \times h} \\$

$\\ \sf \implies \: { 6,900 \: = \: 230 \: \times \: h} \\$

$\\ \sf \implies \: { \: h \: = \: \frac{6900}{230} \: } \\$

$\\ \sf \implies \: { \: h \: = \: 30 \: {cm }^{3} \: } \\$

$\\ \implies{\underline{\boxed{\frak { Height = 30 \: cm^3 }}}} \; \purple\bigstar \: \\$

$\\\qquad \rule{120pt}{3pt}\\$

Therefore , the Volume of the Box is 30 cm ³.

5 0

2 years ago

Use properties to rewrite the given equation. Which equations have the same solution as 3/5x +2/3 + x = 1/2– 1/5x? Check all tha

vodomira [7]

we have

$\frac{3}{5}x+ \frac{2}{3}+x=\frac{1}{2}-\frac{1}{5}x$

Combine like terms in both sides

$(\frac{3}{5}x+ x)+\frac{2}{3}=\frac{1}{2}-\frac{1}{5}x$

we know that

$(\frac{3}{5}x+ x)=(\frac{3}{5}x+ \frac{5}{5}x)=\frac{8}{5}x$

substitute in the expression above

$\frac{8}{5}x+\frac{2}{3}=\frac{1}{2}-\frac{1}{5}x$ -----> equation A

Multiply equation A by $5*3*2=30$ both sides

$30*(\frac{8}{5}x+\frac{2}{3})=30*(\frac{1}{2}-\frac{1}{5}x)$

$48x+20=15-6x$ ---------> equation B

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$48x+6x=15-20$

$54x=-5$ ---------> equation C

Solve for x

$x=-\frac{5}{54} =-0.09$

We are going to proceed to verify each case to determine the solution.

<u>Case a)</u> $\frac{8}{5}x+\frac{2}{3}=\frac{1}{2}-\frac{1}{5}x$

the case a) is equal to the equation A

the case a) have the same solution that the given equation

<u>Case b)</u> $18x+20+30x=15-6x$

Combine like terms in left side

$(18x+30x)+20=15-6x$

$(48x)+20=15-6x$

the case b) is equal to the equation B

the case b) have the same solution that the given equation

<u>Case c)</u> $18x+20+x=15-6x$

Combine like terms in left side

$(18x+x)+20=15-6x$

$(19x)+20=15-6x$

$19x+6x=15-20\\25x=-5\\x=-0.20$

$-0.20\neq -0.09$

therefore

the case c) not have the same solution that the given equation

<u>Case d)</u> $24x+30x=-5$

Combine like terms in left side

$54x=-5$

the case d) is equal to the equation C

the case d) have the same solution that the given equation

<u>Case e)</u> $12x+30x=-5$

Combine like terms in left side

$42x=-5$

$x=-5/42=-0.12$

$-0.12\neq -0.09$

therefore

the case e) not have the same solution that the given equation

therefore

<u>the answer is</u>

case a) $\frac{8}{5}x+\frac{2}{3}=\frac{1}{2}-\frac{1}{5}x$

case b) $18x+20+30x=15-6x$

case d) $24x+30x=-5$

7 0

4 years ago

Read 2 more answers

In a test consisting of 12 questions, each student gets a score of +23 for every correct answer and −13 for every incorrect answ

pentagon [3]

Answer:

96 scores

Step-by-step explanation:

Score for each correct answer = +23

Score for each incorrect answer = $-13$

Each question must be answered such that Ram gets 7 correct and 5 incorrect answers.

Number of correct answers $=7$

Number of incorrect answers $=5$

Therefore,

Ram's score $=7(23)+5(-13)=161-65=96$

4 0

3 years ago

Express the sum of 3x^2+5x-6 and -x^2+3×+9?

djverab [1.8K]

I hope this helps you

6 0

4 years ago

Use calculus to find the absolute maximum and minimum values of the function. (round all answers to three decimal places.) f(x)

Allisa [31]

Part A:

Given the function $f(x)=x+2\cos(x)$ , the absolute maximum or minimum occurs when $f'(x)=0$ .

$f'(x)=0 \\ \\ \Rightarrow1-2\sin{x}=0 \\ \\ \Rightarrow2\sin{x}=1 \\ \\ \Rightarrow\sin{x}= \frac{1}{2} \\ \\ \Rightarrow x=\sin^{-1}{\frac{1}{2}}= \frac{\pi}{6}$

Using the second derivative test,

$f''(x)=-2cosx \\ \\ \Rightarrow f''\left( \frac{\pi}{6} \right)=-2\cos{\left( \frac{\pi}{6} \right)}=-1.732$

Since the second derivative gives a negative number, the given function has a maximum point at $x=\frac{\pi}{6}$ .

And the maximum point is given by:

$f\left( \frac{\pi}{6} \right)=\frac{\pi}{6}+2\cos\left( \frac{\pi}{6} \right) \\ \\ =0.5236+2(0.8660)=0.5236+1.732 \\ \\ =\bold{2.256}$

i.e. $\left(\frac{\pi}{6},\ 2.256\right)$

Part B:

Given the function $f(x)=e^{-x}-e^{-2x}$ , the absolute maximum or minimum occurs when $f'(x)=0$ .

$f'(x)=0 \\ \\ \Rightarrow-e^{-x}+2e^{-2x}=0 \\ \\ \Rightarrow2e^{-2x}=e^{-x} \\ \\ \Rightarrow2e^{-x}=1 \\ \\ \Rightarrow e^{-x}=\frac{1}{2} \\ \\ \Rightarrow-x=\ln \frac{1}{2}=-0.6931 \\ \\ \Rightarrow x=0.6931$

Using the second derivative test,

$f''(x)=e^{-x}-4e^{-2x} \\ \\ \Rightarrow f''(0.6931)=e^{-0.6931}-4e^{-2(0.6931)} \\ \\ =0.5-4e^{-1.386}=0.5-4(0.25)=0.5-1 \\ \\ =-0.5$

Since the second derivative gives a negative number, the given function has a maximum point at $x=0.6931$ .

And the maximum point is given by:

$f(0.6931)=e^{-0.6931}-e^{-2(0.6931)} \\ \\ =0.5-e^{-1.386}=0.5-0.25=\bold{0.25}$

i.e. (0.693, 0.25)

3 0

3 years ago