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3 years ago

Can you guyz help me with this question step by step please ☺️

Mathematics

1 answer:

zlopas [31]3 years ago

7 0

2 1/2 x 3 = 7 1/2 = 7 4/8

10 11/8
- 7 4/8
3 7/8
A is the correct answer

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Which equation represents a line that passes through (–2, 4) and has a slope of ? y – 4 = 2/5(x + 2) y + 4 = 2/5(x – 2) y + 2 =

valentinak56 [21]

$\bf \begin{array}{lllll}
&x_1&y_1\\
% (a,b)
&({{ -2}}\quad ,&{{ 4}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{2}{5}
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-4=\cfrac{2}{5}[x-(-2)]
\\\\\\
y-4=\cfrac{2}{5}(x+2)$

7 0

3 years ago

Write the equation using function notation f(x). then find f(0)<br><br> 2x^2 - 3y = 6

Grace [21]

$\quad \huge \quad \quad \boxed{ \tt \:Answer }$

$\qquad \tt \rightarrow \: f(0) = -2$

____________________________________

$\large \tt Solution \: :$

$\qquad \tt \rightarrow \: 2 {x}^{2} - 3y = 6$

$\qquad \tt \rightarrow \: 3y + 6 = 2 {x}^{2}$

$\qquad \tt \rightarrow \: 3y = 2 {x}^{2} - 6$

$\qquad \tt \rightarrow \: y = \cfrac{2 {x}^{2} - 6 }{3}$

$\qquad \tt \rightarrow \: y = \cfrac{2 {}^{} }{3} {x}^{2} - 2$

[ here, y can be replaced with f(x) because y is a function of x ]

$\qquad \tt \rightarrow \: f(x) = \cfrac{2 {}^{} }{3} {x}^{2} - 2$

$\large\textsf{Find f(0):}$

$\qquad \tt \rightarrow \: f(0) = \cfrac{2 {}^{} }{3} {(0)}^{2} - 2$

$\qquad \tt \rightarrow \: f(0) = 0 - 2$

$\qquad \tt \rightarrow \: f(0) = - 2$

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

8 0

2 years ago

Solve this equation 3x+5-x=x-3

QveST [7]

I think the answer is x=-8
you are basically simplifying both sides of the equation..then isolating the variable

8 0

3 years ago

Explain why 3(x+4)=3(x-5) has no solution. Choose the best response below.

Evgesh-ka [11]

Its either A or C i know that

3 0

3 years ago

Read 2 more answers

An article in Technometrics (Vol. 19, 1977, p. 425) presents the following data on the motor fuel octane ratings of several blen

Slav-nsk [51]

Answer:

$Median = 90.4$

$Q_1 = 88.6$

$Q_3 = 92.2$

Step-by-step explanation:

Given

The above data

Required

- A stem and leaf display

- The median

- The quartiles

First, determine the range of the data

$Smallest = 83.4$

$Highest = 100.3$

Next, group each dataset base on common whole numbers.

So, we have:

$83.4$

$84.3\ 84.3$

$85.3$

$86.7\ 86.7\ 86.7$

$87.4\ 87.5\ 87.6\ 87.7\ 87.8\ 87.9$

$88.2\ 88.3\ 88.3\ 88.3\ 88.4\ 88.5\ 88.5\ 88.6\ 88.6\ 88.7\ 88.9$

$89.0\ 89.2\ 89.3\ 89.3\ 89.6\ 89.7\ 89.8\ 89.8\ 89.9\ 89.9$

$90.0\ 90.1\ 90.1\ 90.1\ 90.3\ 90.4\ 90.4\ 90.4\ 90.5\ 90.6\ 90.7\ 90.8\ 90.9$

$91.0\ 91.0\ 91.0\ 91.1\ 91.1\ 91.1\ 91.2\ 91.2\ 91.2\ \ 91.5\ 91.6\ 91.6\ 91.8\ 91.8$

$92.2\ 92.2\ 92.2\ 92.3\ 92.6\ 92.7\ 92.7\ 92.7$

$93.0\ 93.2\ 93.3\ 93.3\ 93.4\ 93.7$

$94.2\ 94.2\ 94.4\ 94.7$

$96.1\ 96.5$

$98.8$

$100.3$

Next, we construct the stem and leaf plot.

The whole numbers will be the stem while the decimal parts will be the leaf.

So, we have:

$\begin{array}{ccc}{Stem} & {} & {Leaf} & {83} & {|} & {.4} & {84} & {|} & {.3\ .3} & {85} & {|} & {.3} & {86} & {|} & {.7\ .7\ .7} & {87} &{|} & {.4\ .5\ .6\ .7\ .8\ .9} & {88} & {|} & {.2\ .3\ .3\ .3\ .4\ .5\ .5\ .6\ .6\ .7\ .9} &{89} & {|} & {.0\ .2\ .3\ .3\ .6\ .7\ .8\ .8\ .9\ .9} & {90} & {|} &{.0\ .1\ .1\ .1\ .3\ .4\ .4\ .4\ .5\ .6\ .7\ .8\ .9} & {91} &{|}&{.0\ .0\ .0\ .1\ .1\ .1\ .2\ .2\ .2\ .5\ .6\ .6\ .8\ .8} &{92} &{|} &{.2\ .2\ .2\ .3\ .6\ .7\ .7\ .7} \ \end{array}$

$\begin{array}{ccc} {93} & {|} & {.0\ .2\ .3\ .3\ .4\ .7} & {94} &{|} & {.2\ .2\ .4\ .7} &{96} & {|} & {.1\ .5} & {98} & {|} & {.8} & {100} &{|} &{.3} \ \end{array}$