All categories

2 years ago

Which question can be answered by solving the inequality 7x<50?

Mathematics

2 answers:

mote1985 [20]2 years ago

7 0

Answer:

Clayton spent $7 yesterday and now has less than $50. What is the most money, x, he could have had before yesterday?

sdas [7]2 years ago

3 0

Answer:

A. Clayton spent $7 yesterday and now has less than $50. What is the most money, x, he could have had before yesterday?

<em>good luck, i hope this helps :)</em>

You might be interested in

If CDE ~ GDF, find ED

qaws [65]

Answer:

Step-by-step explanation:

$\triangle CDE \sim \triangle GDF.. (given) \\\\\therefore \frac{CD}{GD} =\frac{DE}{DF}.. (csst) \\\\\therefore \frac{15}{x+3} =\frac{3x+1}{4}\\\\ \therefore \: 15 \times 4 = (x + 3)(3x + 1) \\ \\ \therefore \: 60 = 3 {x}^{2} + x + 9x + 3 \\ \\ \therefore 3 {x}^{2} + 10x + 3 - 60 = 0 \\ \therefore 3 {x}^{2} + 10x - 57 = 0 \\ \therefore 3 {x}^{2} + 19x - 9x - 57 = 0 \\ \therefore \: x(3x + 19) - 3(3x + 19) = 0 \\\therefore \: (3x + 19)(x - 3) = 0 \\ \therefore \: 3x + 19 = 0 \: \: or \: \: x - 3 = 0 \\ \therefore \: x = - \frac{19}{3} \: \: or \: \: x = 3 \\ \because \: x \: can \: not \: be \: - ve \\ \therefore \: x = 3 \\ ED = 3x + 1 = 3 \times 3 + 1 \\ \huge \red{ \boxed{ ED= 10}}$

7 0

3 years ago

I really am confused on how to write a proof.

kvasek [131]

Since ABCD is a parallelogram, the opposite sides will be parallel and equal,

$\begin{gathered} AB=CD \\ BC=AD \end{gathered}$

Consider that AC acts as a transversal to the parallel lines AB and CD, so we can write,

$\begin{gathered} \angle CAD=\angle ACB\text{ (Alternate Interior Angles)} \\ BC=AD\text{ (Opposite sides of parallelogram)} \\ \angle ADB=\angle CBD\text{ (Alternate Interior Angles)} \end{gathered}$

So by the ASA criteria, the triangle AED is congruent to the triangle CEB,

Then the corresponding parts of the triangles will be equal,

$\begin{gathered} AE=CE \\ BE=DE \end{gathered}$

Hence Proved.

8 0

1 year ago

PLEASE PLEASE PLEASE I BEG YOUWhich graph is a function of x? On a coordinate plane, a vertical line is at x = negative 3. On a

ki77a [65]

Answer:

On a coordinate plane, a v-shaped graph opens up.

Step-by-step explanation:

If you can draw a vertical line that intersects the graph in more than one place, the graph is NOT a function.

It appears your only function is ...

On a coordinate plane, a v-shaped graph opens up.

4 0

3 years ago

Read 2 more answers

WILL GIVE BRAINLIEST!

Leto [7]

Answer:

I'm not sure how to answer this.

Step-by-step explanation:

Is this the full question? The context seems to be missing something. Who are Ben and Scott and who are the other 2 candidates?

Maybe post another question and rephrase it.

5 0

3 years ago

Read 2 more answers

How do you find a vector that is orthogonal to 5i + 12j ?

Rashid [163]

$\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad \cfrac{a}{b}\\\\
slope=\cfrac{a}{{{ b}}}\qquad negative\implies -\cfrac{a}{{{ b}}}\qquad reciprocal\implies - \cfrac{{{ b}}}{a}\\\\
-------------------------------\\\\$

$\bf \boxed{5i+12j}\implies 
\begin{array}{rllll}
\ \textless \ 5&,&12\ \textgreater \ \\
x&&y
\end{array}\quad slope=\cfrac{y}{x}\implies \cfrac{12}{5}
\\\\\\
slope=\cfrac{12}{{{ 5}}}\qquad negative\implies -\cfrac{12}{{{ 5}}}\qquad reciprocal\implies - \cfrac{{{ 5}}}{12}
\\\\\\
\ \textless \ 12, -5\ \textgreater \ \ or\ \ \textless \ -12,5\ \textgreater \ \implies \boxed{12i-5j\ or\ -12i+5j}$

if we were to place <5, 12> in standard position, so it'd be originating from 0,0, then the rise is 12 and the run is 5.

so any other vector that has a negative reciprocal slope to it, will then be perpendicular or "orthogonal" to it.

so... for example a parallel to <-12, 5> is say hmmm < -144, 60>, if you simplify that fraction, you'd end up with <-12, 5>, since all we did was multiply both coordinates by 12.

or using a unit vector for those above, then

$\bf \textit{unit vector}\qquad \cfrac{\ \textless \ a,b\ \textgreater \ }{||\ \textless \ a,b\ \textgreater \ ||}\implies \cfrac{\ \textless \ a,b\ \textgreater \ }{\sqrt{a^2+b^2}}\implies \cfrac{a}{\sqrt{a^2+b^2}},\cfrac{b}{\sqrt{a^2+b^2}}
\\\\\\
\cfrac{12,-5}{\sqrt{12^2+5^2}}\implies \cfrac{12,-5}{13}\implies \boxed{\cfrac{12}{13}\ ,\ \cfrac{-5}{13}}
\\\\\\
\cfrac{-12,5}{\sqrt{12^2+5^2}}\implies \cfrac{-12,5}{13}\implies \boxed{\cfrac{-12}{13}\ ,\ \cfrac{5}{13}}$

4 0

3 years ago