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3 years ago

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Help! With Fraction Problems??? Hailey is knitting a scarf. Each half hour, she adds 3/7 inch to the scarf's length. How much le

ngth will she add to the scarf in 6 hours?

1 answer:

allsm [11]3 years ago

7 0

18/7 or 2 4/7 bc 3/7 times 6 is 18/7 or 2 and 4/7

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Given:<br> DBC = RST<br><br> Prove:<br> ABC > RST

masya89 [10]

Answer:

<h3><ABC > <DBC.</h3>

Step-by-step explanation:

Given < DBC = < RST and we need to prove < ABC is greater than <RST.

First given statement:

< DBC = < RST

Reason: Given.

Second given statement :

<ABC = <DBC+ <ABD.

Reason: Angle addition theorem.

<em>Note: < ABC is the sum of angles <DBC and <ABD and we have < DBC = < RST. So it's an obvious thing that the sum of angles <DBC and <ABD is always greater than <RST.</em>

Also, <ABC is greater than <DBC.

Therefore, correct option for third statement is :

<h3><ABC > <DBC.</h3>

5 0

3 years ago

If a + 1 = b, then b > a.

kogti [31]

If a=-10;then:
a+1=-10+1=-9=b
-9>-10

if a=10; then
a+1=10+1=11=b
11>10
A)The pair of values for a and be are: a=-10, then the value of b would be: -9;
And a=10; then the value of b would be:11;

B) It isn´t possible to create a pair of values for a and be, in wich the numerical relationship shown in the given conditional stament is false, therefore b>a if a+1=b

4 0

3 years ago

Read 2 more answers

Solve the following equation. show work. x^2 + 10x = -25

SVEN [57.7K]

Step-by-step explanation:

x^2 + 10x = -25

x^2 + 10x + 25 = 0

（x+5)^2=0

x=－5

6 0

2 years ago

Tomorrow, Mrs. Wendel's class will be using toothpicks for a science project. Each student must use at least 5 toothpicks for th

Amiraneli [1.4K]

Answer:

I'm pretty sure A but let me know if I'm wrong

7 0

3 years ago

Square root of 2tanxcosx-tanx=0

kobusy [5.1K]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3242555

——————————

Solve the trigonometric equation:

$\mathsf{\sqrt{2\,tan\,x\,cos\,x}-tan\,x=0}\\\\ \mathsf{\sqrt{2\cdot \dfrac{sin\,x}{cos\,x}\cdot cos\,x}-tan\,x=0}\\\\\\ \mathsf{\sqrt{2\cdot sin\,x}=tan\,x\qquad\quad(i)}$

Restriction for the solution:

$\left\{ \begin{array}{l} \mathsf{sin\,x\ge 0}\\\\ \mathsf{tan\,x\ge 0} \end{array} \right.$

Square both sides of (i):

$\mathsf{(\sqrt{2\cdot sin\,x})^2=(tan\,x)^2}\\\\ \mathsf{2\cdot sin\,x=tan^2\,x}\\\\ \mathsf{2\cdot sin\,x-tan^2\,x=0}\\\\ \mathsf{\dfrac{2\cdot sin\,x\cdot cos^2\,x}{cos^2\,x}-\dfrac{sin^2\,x}{cos^2\,x}=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left(2\,cos^2\,x-sin\,x \right )=0\qquad\quad but~~cos^2 x=1-sin^2 x}$

$\mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\cdot (1-sin^2\,x)-sin\,x \right]=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2-2\,sin^2\,x-sin\,x \right]=0}\\\\\\ \mathsf{-\,\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}\\\\\\ \mathsf{sin\,x\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}$

Let

$\mathsf{sin\,x=t\qquad (0\le t$

So the equation becomes

$\mathsf{t\cdot (2t^2+t-2)=0\qquad\quad (ii)}\\\\ \begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{2t^2+t-2=0} \end{array}$

Solving the quadratic equation:

$\mathsf{2t^2+t-2=0}\quad\longrightarrow\quad\left\{ \begin{array}{l} \mathsf{a=2}\\ \mathsf{b=1}\\ \mathsf{c=-2} \end{array} \right.$

$\mathsf{\Delta=b^2-4ac}\\\\ \mathsf{\Delta=1^2-4\cdot 2\cdot (-2)}\\\\ \mathsf{\Delta=1+16}\\\\ \mathsf{\Delta=17}$

$\mathsf{t=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{2\cdot 2}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{4}}\\\\\\ \begin{array}{rcl} \mathsf{t=\dfrac{-1+\sqrt{17}}{4}}&\textsf{ or }&\mathsf{t=\dfrac{-1-\sqrt{17}}{4}} \end{array}$

You can discard the negative value for t. So the solution for (ii) is

$\begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{t=\dfrac{\sqrt{17}-1}{4}} \end{array}$

Substitute back for t = sin x. Remember the restriction for x:

$\begin{array}{rcl} \mathsf{sin\,x=0}&\textsf{ or }&\mathsf{sin\,x=\dfrac{\sqrt{17}-1}{4}}\\\\ \mathsf{x=0+k\cdot 180^\circ}&\textsf{ or }&\mathsf{x=arcsin\bigg(\dfrac{\sqrt{17}-1}{4}\bigg)+k\cdot 360^\circ}\\\\\\ \mathsf{x=k\cdot 180^\circ}&\textsf{ or }&\mathsf{x=51.33^\circ +k\cdot 360^\circ}\quad\longleftarrow\quad\textsf{solution.} \end{array}$

where k is an integer.

I hope this helps. =)

3 0

3 years ago