All categories

4 years ago

H(x)=4x-7. what is h(-3)

Mathematics

2 answers:

Helen [10]4 years ago

7 0

H(-3)
4(-3)-7
-12-7
-19

Marianna [84]4 years ago

3 0

Your notation is wrong. You cannot have big H and little h at the same time. You need to find h(-3) or H(-3).
I assume you need h(-3), which means that the function is h(x) = 4x - 7.

See picture.

You might be interested in

find the number 15 balls consisting of 5 black, 5 blue nd 5 white balls from a box which has 6 black, 7 blue and 8 white balls (

AleksAgata [21]

Answer:

Step-by-step explanation:

A bag contains 4 blue, 5 white, and 6 green balls. ... There are 5 red and 15 black balls in a box, two are picked up at random

4 0

4 years ago

You answer 32 questions correctly on a 45 question test. You need a score of at least 70% to pass. Do you pass? Explain

Taya2010 [7]

Answer:

Yes

Step-by-step explanation:

32 divided by 45 = .7111111111111111111

7 0

3 years ago

Read 2 more answers

Find the EXACT value of sin(A−B) if cos A = 3/5 where A is in Quadrant IV and cos B = 12/13 where B is in Quadrant IV. Assume al

MissTica

$\bf \textit{Sum and Difference Identities} \\\\ sin(\alpha - \beta)=sin(\alpha)cos(\beta)- cos(\alpha)sin(\beta)$

well, for both angles A and B we're on the IV Quadrant, meaning, the sine is negative, the cosine is positive, likewise, the opposite side is negative and the adjacent side for the angle is positive.

$\bf cos(A)=\cfrac{\stackrel{adjacent}{3}}{\underset{hypotenuse}{5}}\qquad \qquad \stackrel{\textit{getting the opposite side}}{b=\pm\sqrt{5^2-3^2}}\implies b = \pm 4 \\\\\\ \stackrel{IV~Quadrant}{b = -4}\qquad \qquad sin(A)=\cfrac{\stackrel{opposite}{-4}}{\underset{hypotenuse}{5}} \\\\[-0.35em] ~\dotfill\\\\ cos(B)=\cfrac{\stackrel{adjacent}{12}}{\underset{hypotenuse}{13}}\qquad \qquad \stackrel{\textit{getting the opposite side}}{b=\pm\sqrt{13^2-12^2}}\implies b = \pm 5$

$\bf \stackrel{IV~Quadrant}{b = -5}\qquad \qquad sin(B)=\cfrac{\stackrel{opposite}{-5}}{\underset{hypotenuse}{13}} \\\\[-0.35em] ~\dotfill\\\\ sin(A-B)=\cfrac{-4}{5}\cdot \cfrac{12}{13}-\left( \cfrac{3}{5}\cdot \cfrac{-5}{13} \right)\implies sin(A-B)=\cfrac{-48}{65} - \left( \cfrac{-15}{65} \right) \\\\\\ sin(A-B)=\cfrac{-48}{65} + \cfrac{15}{65}\implies sin(A-B)=\cfrac{-33}{65}$

4 0

3 years ago

Harry worked 7 hours last week this is one third as many hours as Aiden work how many hours did Aiden work

Eduardwww [97]

21 hrs. Because if Harry worked one third the hrs thats Aiden worked then 7×3 =21 and 21÷3=7. So 7 is one third of 21.

6 0

3 years ago

Juan owns 7 pairs of pants, 3 shirts, 5 ties, and 6 jackets. How many different outfits can he wear to school if he must wear on

ZanzabumX [31]

To solve this problem, just use multiplication.

$7\text{ pairs of pants}*3\text{ shirts}*5\text{ ties}*6\text{ jackets}=\large\boxed{630\text{ outfits}}$

You can somewhat think of this situation as a sort of tree. You start with 7 large branches representing the pants, then for every one of those you have 3 medium-sized branches to represent the pairs of pants. Each one of those has 5 sticks for the ties, which in turn each have 6 leaves to represent the jackets and therefore a complete outfit. The solution is found by getting the total number of leaves on the tree.

8 0

3 years ago