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3 years ago

13

mae walked a total distance of 25 km in 4 days. she walked 4 2/5 km to first day, 5 km the second day, and 7 3/10 km the third d

ay. How far did Mae walk the last day?

1 answer:

Alika [10]3 years ago

6 0

Mae walked 8 3/10 km on the last day.

If you add 4 2/5+5+7 3/10=16 7/10. Then you subtract 16 7/10 from the total, or 25 km and get 8 3/10 km. Hope this helped:)

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*NO 194 39% 13:08 Pilih jawaban yang paling tepat Segitiga PQR dan segitiga XYZ 5 poin merupakan dua segitiga yang kongruen. Bes

Illusion [34]

Answer:

Option d. PQ = YZ

Step-by-step explanation:

<u><em>The question in English is</em></u>

Choose the most appropriate answer. The PQR triangle and the XYZ triangle are two congruent triangles. The angle P = the angle Y and PR = YX. Side pairs of the same length are ... a. PQ = XZ b. QR = YZ c. QR = XY d. PQ = YZ

we know that

If two triangles are congruent, then its corresponding sides and its corresponding angles are congruent

Corresponding sides are named using pairs of letters in the same position on either side of the congruence statement

so

we have that

$m\angle P=m\angle Y$

$PR=YX$

so

Triangle PQR is congruent with Triangle YZX

That means

<em><u>Corresponding angles</u></em>

$m\angle P=m\angle Y$

$m\angle Q=m\angle Z$

$m\angle R=m\angle X$

<u><em>Corresponding sides</em></u>

$PQ=YZ\\QR=ZX\\PR=YX$

8 0

3 years ago

A soccer player stands in the south east corner of a 72 yard x 65 yard field and kick the soccer ball All the way to a player in

lubasha [3.4K]

Answer:

Using the Pythagorean Theorem:

distance^2 = 72^2 + 65^2

distance^2 = 5,184 + 4,225

distance^2 = 9,409

distance = 97 yards

"round-trip" distance = 97 * 2 = 194 yards

Step-by-step explanation:

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3 years ago

How do you calculate the area of triangle DEF?

cricket20 [7]

First you need to solve for angle F then you can use 1/2acSinB. If you haven't done this you can also use the Pythagoras theorem and solve for the unknown side once you do that you can just use 1/2 b*h

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3 years ago

Read 2 more answers

For 0 ≤ ϴ < 2π, how many solutions are there to tan(StartFraction theta Over 2 EndFraction) = sin(ϴ)? Note: Do not include va

Black_prince [1.1K]

Answer:

3 solutions:

$\theta={0, \frac{\pi}{2}, \frac{3\pi}{2}}$

Step-by-step explanation:

So, first of all, we need to figure the angles that cannot be included in our answers out. The only function in the equation that isn't defined for some angles is $tan(\frac{\theta}{2})$ so let's focus on that part of the equation first.

We know that:

$tan(\frac{\theta}{2})=\frac{sin(\frac{\theta}{2})}{cos(\frac{\theta}{2})}$

therefore:

$cos(\frac{\theta}{2})\neq0$

so we need to find the angles that will make the cos function equal to zero. So we get:

$cos(\frac{\theta}{2})=0$

$\frac{\theta}{2}=cos^{-1}(0)$

$\frac{\theta}{2}=\frac{\pi}{2}+\pi n$

or

$\theta=\pi+2\pi n$

we can now start plugging values in for n:

$\theta=\pi+2\pi (0)=\pi$

if we plugged any value greater than 0, we would end up with an angle that is greater than $2\pi$ so, that's the only angle we cannot include in our answer set, so:

$\theta\neq \pi$

having said this, we can now start solving the equation:

$tan(\frac{\theta}{2})=sin(\theta)$

we can start solving this equation by using the half angle formula, such a formula tells us the following:

$tan(\frac{\theta}{2})=\frac{1-cos(\theta)}{sin(\theta)}$

so we can substitute it into our equation:

$\frac{1-cos(\theta)}{sin(\theta)}=sin(\theta)$

we can now multiply both sides of the equation by $sin(\theta)$

so we get:

$1-cos(\theta)=sin^{2}(\theta)$

we can use the pythagorean identity to rewrite $sin^{2}(\theta)$ in terms of cos:

$sin^{2}(\theta)=1-cos^{2}(\theta)$

so we get:

$1-cos(\theta)=1-cos^{2}(\theta)$

we can subtract a 1 from both sides of the equation so we end up with:

$-cos(\theta)=-cos^{2}(\theta)$

and we can now add $cos^{2}(\theta)$

to both sides of the equation so we get:

$cos^{2}(\theta)-cos(\theta)=0$

and we can solve this equation by factoring. We can factor $cos(\theta)$ to get:

$cos(\theta)(cos(\theta)-1)=0$

and we can use the zero product property to solve this, so we get two equations:

Equation 1:

$cos(\theta)=0$

$\theta=cos^{-1}(0)$

$\theta={\frac{\pi}{2}, \frac{3\pi}{2}}$

Equation 2:

$cos(\theta)-1=0$

we add a 1 to both sides of the equation so we get:

$cos(\theta)=1$

$\theta=cos^{-1}(1)$

$\theta=0$

so we end up with three answers to this equation:

$\theta={0, \frac{\pi}{2}, \frac{3\pi}{2}}$

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2 years ago

Which of the following functions f : {0, 1, 2, 3} ! {0, 1, . . . , 7} are one-to-one?

allsm [11]

Answer:

1. No.

$f(0)=0\\1^2=1; \text{then } f(1)=1\\2^2=4\equiv 4 \text{ mod 8}; \text{then } f(2)=4\\3^2=9\equiv 1 \text{mod 8 }; \text{then } f(3)=1$

Since f(1)=f(3) and $1\neq 3$ then f isn't one-to-one.

2. No

$f(0)=0\\1^3=1\equiv 1\text{ mod 8}; \text{then } f(1)=1\\2^3=8\equiv 0\text{ mod 8}; \text{then } f(2)=0\\3^3=27\equiv 3 \text{ mod 8}; \text{then } f(3)=3$

Since f(0)=f(2) and $0\neq 2$ then f isn't one-to-one.

3. No

$0^3-8=-8\equiv 0\text{ mod 8}; \text{then } f(0)=0\\1^3-8=-7\equiv 1\text{ mod 8}; \text{then } f(1)=1\\2^3-8=0\equiv 0 \text{ mod 8}; \text{then } f(2)=0\\3^3-8=27-8=19\equiv 3 \text{ mod 8}; \text{then } f(3)=3\\$

Since f(0)=f(2) and $0\neq 2$ then f isn't one-to-one.

4. Yes

$0^3+2*0=0; \text{then } f(0)=0\\1^3+2*1=3\equiv 3\text{ mod 8}; \text{then } f(1)=3\\2^3+2*2=8+4=12\equiv 4 \text{ mod 8}; \text{then } f(2)=4\\3^3+2*3=27+6=33\equiv 1\text{ mod 8}; \text{then } f(3)=1$

Since $f(0)\neq f(1)\neq f(2) \neq f(3)$ , then f is one-to-one

5. Since f(1)=f(3) and $1\neq 3$ then, f isn't one-to-one

3 0

3 years ago