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3 years ago

Five less than the product of 3 and a number of 40 . Write in an equation

Mathematics

2 answers:

Serjik [45]3 years ago

8 0

Answer:

(3*40)-5=115

Step-by-step explanation:

vagabundo [1.1K]3 years ago

4 0

Answer:

I don't get the "number of 40" part, but here's my best shot!

3(40x)-5

Step-by-step explanation:

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Factorise x squared + 4x + 3

olganol [36]

<h3>Answer: (x+1)(x+3)</h3>

===================================================

Explanation:

Let's assume it factors into (x+a)(x+b)

The goal is to find the two numbers a and b.

FOIL out (x+a)(x+b) to get x^2+(a+b)x+ab

Note how a+b is the middle term and ab is the last term.

In the original expression, 4 is the middle term and 3 is the last term.

So we need to find two numbers that

add to 4
multiply to 3

There are two ways to multiply to 3 and they are

1 times 3 = 3
-1 times -3 = -3

But only the first way has the factors add to 4. So that means a = 1 and b = 3.

Therefore (x+a)(x+b) = (x+1)(x+3)

And x^2+4x+3 = (x+1)(x+3)

8 0

3 years ago

Work out the value of (2^3)^2

aniked [119]

Answer:

2^6

Step-by-step explanation:

7 0

3 years ago

How many minutes are in a 2 days?

Fantom [35]

There are 2880 minutes in two days

3 0

3 years ago

Read 2 more answers

Area of a triangle with points at (-9,5), (6,10), and (2,-10)

Ann [662]

First we are going to draw the triangle using the given coordinates.
Next, we are going to use the distance formula to find the sides of our triangle.
Distance formula: $d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Distance from point A to point B:
$d_{AB}= \sqrt{[6-(-9)]^2+(10-5)^2}$
$d_{AB}= \sqrt{(6+9)^2+(10-5)^2}$
$d_{AB}= \sqrt{(15)^2+(5)^2}$
$d_{AB}= \sqrt{225+25}$
$d_{AB}= \sqrt{250}$
$d_{AB}=15.81$

Distance from point A to point C:
$d_{AC}= \sqrt{[2-(-9)]^2+(-10-5)^2}$
$d_{AC}= \sqrt{(2+9)^2+(-10-5)^2}$
$d_{AC}= \sqrt{11^2+(-15)^2}$
$d_{AC}= \sqrt{121+225}$
$d_{AC}= \sqrt{346}$
$d_{AC}= 18.60$

Distance from point B from point C
$d_{BC}= \sqrt{(2-6)^2+(-10-10)^2}$
$d_{BC}= \sqrt{(-4)^2+(-20)^2}$
$d_{BC}= \sqrt{16+400}$
$d_{BC}= \sqrt{416}$
$d_{BC}=20.40$

Now, we are going to find the semi-perimeter of our triangle using the semi-perimeter formula:
$s= \frac{AB+AC+BC}{2}$
$s= \frac{15.81+18.60+20.40}{2}$
$s= \frac{54.81}{2}$
$s=27.41$

Finally, to find the area of our triangle, we are going to use Heron's formula:
$A= \sqrt{s(s-AB)(s-AC)(s-BC)}$
$A=\sqrt{27.41(27.41-15.81)(27.41-18.60)(27.41-20.40)}$
$A= \sqrt{27.41(11.6)(8.81)(7.01)}$
$A=140.13$

We can conclude that the perimeter of our triangle is 140.13 square units.

3 0

3 years ago

2/3q = 1/2 what does q equal as a decimal

goldfiish [28.3K]

Is that 2/3 of q or 2 over 3q?
for 2/3 of q. q=1/2 x 3/2 =3/4=0.75

for 2 over 3q q = 2 x 2/3= 4/3=1.3333

3 0

3 years ago