Please help me on this moby max question

Mathematics

1 answer:

pantera1 [17]3 years ago

7 0

The answer would be 49 + 42b if you multiply 7*7 and 7*6b

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Solve:<br> 4 - 10x >-21<br> and aranh the

Luden [163]

Answer:

$x < \frac{5}{2}$

Step-by-step explanation:

$4 - 10x > -21$

$-10x > -25\\\\x > \frac{-25}{-10} \\\\x > \frac{5}{2}$

8 0

3 years ago

Given: ABC is a right triangle with right angle C. AC = 15 centimeters and mZA= 40°

kupik [55]

Answer:

hello luv

Step-by-step explanation:

i see ur struggling mate

6 0

3 years ago

If EG=67, then EF=?<br> There is a picture below

ikadub [295]

If EG=67 we have (3x+7)+(5x-4)=67
let's simplify
$67 = (3x + 7) + (5x - 4) \\ 67 = 8x + 3 \\ 64 = 8x \\ \frac{64}{8} = x \\ 8 = x$
now we know x is 8 we can use this to find EF
since EF=3x+7
$3(8) + 7 = 24 + 7 = 31$
so we have EF is 31!
if.you like my answer please rate Brainliest

3 0

3 years ago

Read 2 more answers

Can someone PLEASE help me with Trigonometry???

Svetach [21]

$\bf sin^2(2\theta)-7sin(2\theta)-1=0\\\\
-------------------------------\\\\
sin(2\theta)=\cfrac{7\pm\sqrt{49-4(1)(-1)}}{2(1)}\implies sin(2\theta)=\cfrac{7\pm\sqrt{49+4}}{2}
\\\\\\
sin(2\theta)=\cfrac{7\pm\sqrt{53}}{2}\implies 2\theta=sin^{-1}\left( \frac{7\pm\sqrt{53}}{2} \right)
\\\\\\
\theta=\cfrac{sin^{-1}\left( \frac{7\pm\sqrt{53}}{2} \right)}{2}$

but anyway, the numerator will give the angles, and θ is just half of each

$\bf \theta=\cfrac{sin^{-1}\left( \frac{7\pm\sqrt{53}}{2} \right)}{2}\\\\
-------------------------------\\\\
sin^{-1}\left( \frac{7+\sqrt{53}}{2} \right)\implies sin^{-1}(7.14)\impliedby \textit{greater than 1, no good}
\\\\\\
sin^{-1}\left( \frac{7-\sqrt{53}}{2} \right)\implies sin^{-1}(-0.14) \approx -8.05^o
\\\\\\
\theta=\cfrac{-8.05}{2}\implies \theta=-4.025$

ok... that's a negative tiny angle, is in the 4th quadrant, if we stick to the range given, from 0 to 360, so we have to use the positive version of it, 360-4.025

so the angle is 355.975°

now, the 3rd quadrant has another angle whose sine is negative, so... if we move from the 180° line down by 4.025, we end up at 184.025°

and those are the only two angles, because, on the 2nd and 1st quadrants, the sine is positive, so it wouldn't have an angle there

7 0

3 years ago

Two cyclists leave a city at the same time, one going east and the other going west. The westbound cyclist bikes at 3 mph faster

NikAS [45]

We will say that the eastbound cyclist is going x mph. This means that the westbound cyclist is going x+3 mph.
They are both cycling for 6 hours and end up 162 miles apart.
Distance=rate*time
6x+6(x+3)=162
6x+6x+18+162
12x=144
x=12
So, the eastbound cyclist is going 12 mph.
Remember the westbound cyclist is going 3 mph faster, so s/he is going 15 mph.
Hope this helps!!

7 0

3 years ago