All categories

3 years ago

The solution to 4x - 11 = 33 is also a solution of which of the following equations?

Mathematics

1 answer:

Ronch [10]3 years ago

5 0

X= 11 so whatever other equation that has x= 11 would be the answer but I can't see the others...

You might be interested in

(2a-3)(b-a)-3a(3a+b)

ozzi

Answer: -ab - 11a^2 - 3b + 3a

Step-by-step explanation:

i think this is what you're wanting? if not just comment what you need and i'll retry it!

3 0

3 years ago

What % of 400 = 193

saveliy_v [14]

The best way to find your answer is to divide 193 by 400 to get .4825 and to make sure if it is the answer you take 400 and multiply .4825 and it will be 193 so .4825 = 48.25%

3 0

3 years ago

Venessa has scored 45 32 and 37 on her three math quizez she will take one more quiz and she wants a quiz average of at least 40

madreJ [45]

$\frac{45 + 32 + 37 + x}{4} = 40 \\ x = 46$

6 0

3 years ago

An isosceles triangle ABC has legs of length 24 and a vertex angle that measures 36º . Determine the length of its base, BC , to

iris [78.8K]

Answer: $BC=14.8$

Step-by-step explanation:

By definition, an Isosceles triangle has two equal sides and its opposite angles are congruent.

Observe the figure attached, where the isosceles triangle is divided into two equal right triangles.

So, in this case you need to use the following Trigonometric Identity:

$sin\alpha =\frac{opposite}{hypotenuse}$

In this case, you can identify that:

$\alpha =\angle BAD=\angle CAD=18\°\\\\opposite=BD=CD=x\\\\hypotenuse=AB=AC=24$

Substituting values, and solving for "x", you get:

$sin(18\°)=\frac{x}{24}\\\\24*sin(18\°)=x\\\\x=7.416$

Therefore, the length of BC rounded to nearest tenth, is:

$BC=2(7.416)=14.8$

6 0

3 years ago

Anyone have the answer for this

sergiy2304 [10]

$\huge\bold{Given:}$

Length of the base = 16 km.

Length of the hypotenuse = 34 km. $\huge\bold{To\:find:}$

✎ The length of the missing leg '' $a$ ".

$\large\mathfrak{{\pmb{\underline{\orange{Solution}}{\orange{:}}}}}$

The length of the missing leg "a" is $\boxed{30\:km}$ .

$\large\mathfrak{{\pmb{\underline{\red{Step-by-step\:explanation}}{\orange{:}}}}}$

Using Pythagoras theorem, we have

$({perpendicular})^{2} + ({base})^{2} = ({hypotenuse})^{2} \\ ⇢ {a}^{2} + ({16 \: km})^{2} = ({34 \: km})^{2} \\ ⇢ {a}^{2} + 256 \: {km}^{2} = 1156 \: {km}^{2} \\ ⇢ {a}^{2} = 1156 \: {km}^{2} - 256 \: {km}^{2} \\ ⇢ {a}^{2} = 900 \: {km}^{2} \\ ⇢a \: = \sqrt{900 \: {km}^{2} } \\ ⇢a = \sqrt{30 \times 30 \: {km}^{2} } \\ ⇢a = 30 \: km$

$\sf\blue{Therefore,\:the\:length\:of\:the\:missing\:leg\:"a"\:is\:30\:km.}$

$\huge\bold{To\:verify :}$

$( {30 \: km})^{2} + ({16 \: km})^{2} =( {34 \: km})^{2} \\ ⇝900 \: {km}^{2} + 256 \: {km}^{2} = 1156 \: {km}^{2} \\⇝1156 \: {km}^{2} = 1156 \: {km}^{2} \\ ⇝L.H.S.=R. H. S$

Hence verified. ✔

$\circ \: \: { \underline{ \boxed{ \sf{ \color{green}{Happy\:learning.}}}}}∘$

7 0

3 years ago