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3 years ago

64 w^2 + 96w+36=0. What is the number of real solutions?

Mathematics

2 answers:

patriot [66]3 years ago

5 0

Answer:

1 solution multiplicity 2.

Step-by-step explanation:

64 w^2 + 96w+36=0 Divide through by 4:

16w^2 + 24w + 9 = 0

The value of the discriminant = b^2 - 4ac

= 24^2 - 4*16*9

= 0.

This means there is only 1 real solution of multiplicity 2.

The graph will just touch the x - axis at one point

7nadin3 [17]3 years ago

4 0

Answer:

w = -3/4 or -0.75

there is only one solution

Step-by-step explanation:

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PLEASE HELP ME WITH THIS MATH!!!!!!!!!. pLeAsE i ReAlLy NeEd SoMe HeLp

KonstantinChe [14]

Part 1) we know that
m∠5=44° m∠11=86°
m∠2=m∠5------> by vertical angles
m∠2=44°
m∠13=m∠11------> by vertical angles
m∠13=86°
m∠12+m∠13=180°-----> supplementary angles
m∠12=180-86-----> m∠12=94°
m∠14=m∠12----> by vertical angles
m∠14=94°
m∠1=m∠11----> by corresponding angles
m∠1=86°
m∠4=m∠1----> by vertical angles
m∠4=86°
m∠2+m∠1+m∠6=180
m∠6=180-(86+44)----> 50°
m∠6=50°
m∠3=m∠6----> by vertical angles
m∠3=50°
m∠8=m∠3----> by corresponding angles
m∠8=50°
m∠8+m∠7=180°-----> supplementary angles
m∠7=180-50----> 130°
m∠7=130°
m∠10=m∠6----> by corresponding angles
m∠10=50°
m∠10+m∠9=180°-----> supplementary angles
m∠9=180-50-----> 130°
m∠9=130°

the answers Part 1) are
m∠1=86°
m∠2=44°
m∠3=50°
m∠4=86°
m∠5=44°
m∠6=50°
m∠7=130°
m∠8=50°
m∠9=130°
m∠10=50°
m∠11=86°
m∠12=94°
m∠13=86°
m∠14=94°

Part 2)
a) what is m∠TPR?
in the right triangle PTR
m∠PTR+m∠TPR+m∠TRP=180° ( the sum of internal angles of triangle is equal to 180 degrees)
m∠PTR=30°
m∠TRP=90°
so
m∠TPR=180-(90+30)----> 60°

the answer Part 2a) is
m∠TPR=60°

b) what is the length in inches of segment PR?
in the right triangle PTR
sin 30=PR/TP-----> PR=TP*sin 30-----> PR=14*(1/2)----> 7 in

the answer Part 2b) is
PR=7 in

c) what is the length in inches of segment TR?
in the right triangle PTR
cos 30=TR/PT-----> TR=PT*cos 30-----> TR=14*(√3/2)---> TR=7√3 in

the answer Part 2c) is
TR=7√3 in

d) what is the length in inches of segment PQ?
in the right triangle PQR
PR=7 in
RQ=PR-----> by angle 45°
so
RQ=7 in
applying the Pythagoras Theorem
PQ²=RQ²+PR²-----> 7²+7²-----> PQ²=98-----> PQ=√98 in---> PQ=7√2 in

the answer Part 2d) is
PQ=7√2 in

Part 3) Patrice buys a block of wax in the shape of a right rectangular prism. The dimensions of the block are 20 cm by 9 cm by 8 cm.
 (a) What is the volume of the block?

volume of the prism=20*9*8-----> 1440 cm³

the answer Part 3 a) is
the volume of the block is 1440 cm³

Patrice melts the wax and creates a candle in the shape of a circular cylinder that has a diameter of 10 cm and a height of 15 cm.(b) To the nearest centimeter, what is the volume of the candle?
volume of a cylinder=pi*r²*h
diameter=10 cm
radius r=10/2----> 5 cm
h=15 cm
volume of a cylinder=pi*5²*15----> 1177.5 cm³-----> 1178 cm³

the answer Part 3b) is
the volume of the candle is 1178 cm³

Patrice decides to use the remaining wax to create a candle in the shape of a cube.(c) To the nearest centimeter, what is the length of the side of the cube?

the remaining wax=volume of the prism-volume of a cylinder
=1440-1178-----> 262 cm³

volume of a cube=b³
where b is the length side of the cube
262=b³-------b=∛262-----> b=6.40 cm-----> b=6 cm

the answer Part 3c) is
the length of the side of the cube is 6 cm

5 0

3 years ago

(10 pts) (a) (2 pts) What is the difference between an ordinary differential equation and an initial value problem? (b) (2 pts)

laiz [17]

Answer:

Step-by-step explanation:

(A) The difference between an ordinary differential equation and an initial value problem is that an initial value problem is a differential equation which has condition(s) for optimization, such as a given value of the function at some point in the domain.

(B) The difference between a particular solution and a general solution to an equation is that a particular solution is any specific figure that can satisfy the equation while a general solution is a statement that comprises all particular solutions of the equation.

(C) Example of a second order linear ODE:

M(t)Y"(t) + N(t)Y'(t) + O(t)Y(t) = K(t)

The equation will be homogeneous if K(t)=0 and heterogeneous if $K(t)\neq 0$

Example of a second order nonlinear ODE:

$Y=-3K(Y){2}$

(D) Example of a nonlinear fourth order ODE:

$K^4(x) - \beta f [x, k(x)] = 0$

4 0

3 years ago

I need help with premetier and area for 7 8 9

SCORPION-xisa [38]

The answers to the three shapes are:

Figure 7; perimeter = 16.2mm, area = 12.64 square mm

Figure 8; perimeter = 15.2inches, area = 9.61 square inches

Figure 9; perimeter = 21.47yards, area = 16.81 square yards

<h3>What is the perimeter of a shape?</h3>

Perimeter is the outside boundary of a plane shape.

Analysis:

for figure 7, perimeter = s + s + s = 3s = 3(5.4) = 16.2mm

Height of triangle = $\sqrt{((5.4)^{2} - (2.7)^{2} }$ = 4.68

Area of triangle = 1/2 base x height = 1/2 x 5.4 x 4.68 = 12.64 $mm^{2}$

For figure 8, perimeter = b + b + a = 5.9 + 5.9 + 3.4 = 15.2 inches

Height of triangle = $\sqrt{(5.9)^{2} - (1.7)^{2} }$ = 5.65inches

Area of triangle = 1/2 base x height = 1/2 x 3.4 x 5.65 = 9.61 $inches^{2}$

For figure 9, perimeter = b + a + c = 8.2 + 4.1 + 9.17 = 21.47yards

Area of triangle = 1/2 x base x height = 1/2 x 8.2 x 4.1 = 16.81 $yards^{2}$

In conclusion, the perimeter and area of the given shapes are:

Figure 7; perimeter = 16.2mm, area = 12.64 square mm

Figure 8; perimeter = 15.2inches, area = 9.61 square inches

Figure 9; perimeter = 21.47yards, area = 16.81 square yards

Learn more about perimeter of plane shapes: brainly.com/question/2569205

#SPJ1

5 0

2 years ago

Mrs. Montgomery brought cookies to her class. Each student can eat

seropon [69]

B, because 24/4=6 :))))))

5 0

3 years ago

Determine the value of k

KATRIN_1 [288]

Answer:

$\displaystyle k = 6$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality

Algebra I

Functions
Function Notation

Algebra II

Piecewise Functions

Calculus

Limits
Continuity

Step-by-step explanation:

Step 1: Define

Identify

Continuous at x = 2

$\displaystyle f(x) = \left \{ {{2x^2 \ if \ x < 2} \atop {x + k \ if \ x \geq 2}} \right.$

Step 2: Solve for k

Definition of Continuity: $\displaystyle \lim_{x \to 2^+} 2x^2 = \lim_{x \to 2^-} x + k$
Evaluate limits: $\displaystyle 2(2)^2 = 2 + k$
Evaluate exponents: $\displaystyle 2(4) = 2 + k$
Multiply: $\displaystyle 8 = 2 + k$
[Subtraction Property of Equality] Subtract 2 on both sides: $\displaystyle 6 = k$
Rewrite: $\displaystyle k = 6$

Topic: AP Calculus AB/BC (Calculus I/II)

Unit: Limits - Continuity

Book: College Calculus 10e

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3 years ago

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