All categories

3 years ago

4 times the difference between Evan's age "x," and 44 is 64. Write an equation to express Evan's age.

Mathematics

1 answer:

Aneli [31]3 years ago

7 0

44+64=4X we don’t know Evans age so we have to multiply it by the difference

You might be interested in

2 x 10 in standard form

lilavasa [31]

Answer:

y = 3 suh slid este ob cine este a, b și g au nevoie de acel kinfo pentru a termina

Step-by-step explanation:

bah de fort

3 0

3 years ago

Which statement correctly compares the shapes of the distribution?

aniked [119]

the answer to your question is c

4 0

3 years ago

Look at the figure shown below.

Sveta_85 [38]

Answer:

Step by step explanation:

7 0

3 years ago

The factorization of x2 + 3x – 4 is modeled with algebra

meriva

Option C: (x+4) and (x-1) are the factors of the equation

Explanation:

Given that the equation is $x^{2} +3x-4$

We need to determine the factors of the equation.

Splitting the middle term, we get,

$x^{2} +4x-x-4$

Grouping the terms, we have,

$x(x+4)-1(x+4)$

Let us factor out the term (x+4), we have,

$(x+4)(x-1)$

Thus, the factors of the equation are (x+4) and (x-1)

Hence, Option C is the correct answer.

4 0

4 years ago

Three security cameras were mounted at the corners of a triangles parking lot. Camera 1 was 110 ft from camera 2, which was 137

Nata [24]

Answer:

<em>Camera 2nd has to cover the maximum angle, i.e. </em> $78.70^\circ$ .

Step-by-step explanation:

Please have a look at the triangular park represented as a triangle $\triangle ABC$ with sides

a = 110 ft

b = 158 ft

c = 137 ft

1st camera is located at point C, 2nd camera at point B and 3rd camera at point A respectively.

We can use law of cosines here, to find out the angles $\angle A, \angle B, \angle C$

As per Law of cosine:

$cos C = \dfrac{a^{2}+b^2-c^2 }{2ab}\\cos B = \dfrac{a^{2}+c^2-b^2 }{2ac}\\cos A = \dfrac{b^{2}+c^2-a^2 }{2bc}$

Putting the values of a,b and c to find out angles $\angle A, \angle B, \angle C$ .

$cos C = \dfrac{110^{2}+158^2-137^2 }{2\times 110 \times 158}\\\Rightarrow cos C = \dfrac{12100+24964-18769 }{24760}\\\Rightarrow cos C =0.526\\\Rightarrow C = 58.24^\circ$

$cos B = \dfrac{110^{2}+137^2-158^2 }{2\times 110 \times 137}\\\Rightarrow cos B = \dfrac{12100+18769 -24964}{30140}\\\Rightarrow cos B = \dfrac{5905}{30140}\\\Rightarrow cos B =0.196\\\Rightarrow B = 78.70^\circ$

$cos A = \dfrac{158^{2}+137^2-110^2 }{2\times 158 \times 137}\\\Rightarrow cos A = \dfrac{24964+18769-12100}{43292}\\\Rightarrow cos A = \dfrac{31633}{43292}\\\Rightarrow cos A = 0.731\\\Rightarrow A = 43.05^\circ$

<em>Camera 2nd has to cover the maximum angle</em>, i.e. $78.70^\circ$ .

6 0

4 years ago