All categories

3 years ago

What is the missing value in the image above? *

Mathematics

1 answer:

Ne4ueva [31]3 years ago

5 0

Answer:

Step-by-step explanation:

because there is nothing to subtract 8 from

You might be interested in

_______ are the infections protein particles.<br>

vaieri [72.5K]

Answer:

The answer is prions

6 0

3 years ago

Read 2 more answers

What is the result when the number 35 is increased by 20%?<br> Show me your work!

nirvana33 [79]

Answer:

Step-by-step explanation:

first we calculate what is 20% of 35 is then we add it to 35 to find the increased number

35 ÷ 100 = 0.35 this multipled by 20 = 7

35 + 7 = 42

3 0

3 years ago

Complete factorization f(x)=x^4-x^3-5x^2-x-6

Marianna [84]

The solution to the above factorization problem is given as f′(x)=4x³−3x²−10x−1. See steps below.

<h3>What are the steps to the above answer?</h3>

Step 1 - Take the derivative of both sides

f′(x)=d/dx(x^4−x^3−5x^2−x−6)

Step 2 - Use differentiation rule d/dx(f(x)±g(x))=d/dx(f(x))±d/dx(g(x))

f′(x)=d/dx(x4)−d/dx(x^3)−d/dx(5x^2)−d/dx(x)−d/dx(6)

f′(x)=4x^3−d/dx(x3)−d/dx(5x^2)−d/dx(x)−d/dx(6)

f′(x)=4x^3−3x2−d/dx(5x^2)−d/dx(x)−d/dx(6)

f′(x)=4x^3−3x^2−10x−d/dx(x)−d/dx(6)

f′(x)=4x^3−3x^2−10x−1−dxd(6)

f′(x)=4x^3−3x^2−10x−1−0

Learn more about factorization:
brainly.com/question/25829061
#SPJ1

5 0

2 years ago

There is a triangle with a perimeter of 63 cm, one side of which is 21 cm. Also, one of the medians is perpendicular to one of t

NemiM [27]

Answer:

21cm; 28cm; 14cm

Step-by-step explanation:

There is no info in the problem/s text which one of the triangle's side is 21 cm. That is why we have to try all possible variants.

Let the triangle is ABC . Let the AK is the angle A bisector and BM is median.

Let O is AK and BM cross point.

Have a look to triangle ABM. AO is the bisector and AOB=AOM=90 degrees (means that AO is as bisector as altitude)

=> triangle ABM is isosceles => AB=AM (1)

1. Let AC=21 So AM=21/2=10.5 cm

So AB=10.5 cm as well. So BC= P-AB-AC=63-21-10.5=31.5 cm

Such triangle doesn' t exist ( is impossible), because the triangle's inequality doesn't fulfill. AB+AC>BC ( We have 21+10.5=31.5 => AB+AC=BC)

2. Let AB=21 So AM=21 and AC=42 .So BC= P-AB-AC=63-21-42=0 cm- such triangle doesn't exist.

3. Finally let BC=21 cm. So AB+AC= 63-21=42 cm

We know (1) that AB=AM so AC=2*AB. So AB+AC=AB+2*AB=3*AB

=>3*AB=42=> AB=14 cm => AC=2*14=28 cm.

Let check if this triangle exists ( if the triangle's inequality fulfills).

BC+AB>AC 21+14>28 - correct=> the triangle with the sides' length 21cm,14 cm, 28cm exists.

This variant is the only possible solution of the given problem.

6 0

3 years ago

A circle has centre (3,0) and radius 5. The line y = 2x + k intersects the circle in two points. Find the set

lara [203]

A circle with center $(3,0)$ and radius $5$ has equation

$(x-3)^2+y^2=25 \iff x^2 + y^2- 6 x = 16$

If we substitute $y=2x+k$ in this equation, we have

$x^2+(2x+k)^2-6x=16 \iff 5x^2+(4k-6)x+k^2-16=0$

This equation has two solutions (i.e. the line intersects the circle in two points) if and only if the determinant is greater than zero:

$\Delta=b^2-4ac=(4k-6)^2-4\cdot 5\cdot (k^2-16)>0$

The expression simplifies to

$-4 (k^2 + 12 k - 89)>0 \iff k^2 + 12 k - 89$

The solutions to the associated equation are

$k^2 + 12 k - 89=0 \iff k=-6\pm 5\sqrt{5}$

So, the parabola is negative between the two solutions:

$-6-5\sqrt{5}$

8 0

3 years ago